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3025
Subject
Health Science
Date
Apr 3, 2024
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42
Uploaded by CaptainAardvarkPerson215
Exam 1 Practice Questions for QMB 3250 Data from the following sources were used in writing this set of practice questions. Chen, Jing et al. “Vitamin D Status among the Elderly Chinese Population: A Cross-Sectional Analysis of the 2010–2013 China National Nutrition and Health Survey (CNNHS).” Nutrition Journal
16 (2017): 3. PMC
. Web. 31 Aug. 2017. Cohen, Cathy J., Matthew Fowler, Matthew D. Luttig, Vladimir E. Medenica, and Jon C. Rogowski. 2017. “Education in America: The Views of Millennials”. Questions 1 – 2 Read the below excerpt from the article by Jong Chen and answer the questions below. “In the present study, the prevalence of vitamin D deficiency and insufficiency in a nationally representative sample of the Chinese population over 60 years old was evaluated. . . .Data for the present analysis were extracted from the China National Nutrition and Health Survey (CNNHS), 2010–
2013. This survey was a nationally representative cross-sectional study conducted by the Chinese Center for Disease Control and Prevention to assess the health and nutrition of Chinese civilians. The survey randomly selected 150 districts (urban) or counties (rural) within all thirty-one provinces, autonomous regions and municipalities directly under the central Chinese government (except Taiwan, Hong Kong and Macao). “ 1.
What is the response variable? a)
Location (urban or rural) b)
Socioeconomic Status c)
Vitamin D deficiency/insufficiency d)
Age C This study is trying to determine the prevalence of Vitamin D deficiency. 2.
To look at the relationship between Serum 25(OD)D levels and location(urban/rural), what type of graph might you use? a)
pie chart b)
scatterplot c)
boxplots d)
bar charts C Serum 25 Od(d) levels is a quantitative variable and location is a categorical variable, so you would use a boxplot plot to examine a categorical versus
Questions 3 - 5 The World Bank defines the poverty line to be $1.90 a day. The value of $1.90 represents the international equivalent of what $1.90 could buy you in the US in 2011. Below is a scatterplot of the percent of people in Brazil leaving on less than $1.90 a day versus year.
(Data from the World Bank.) 3. Interpret the slope (if applicable). a)
Each year, the percent below the poverty line in Brazil decreases by 0.6783%. b)
Each year, the percent below the poverty line in Brazil tends to decrease by 0.6783%. c)
Each year, the percent below the poverty line in Brazil tends to decrease by 3.26%. d)
0.6783% of variability of Brazilian poverty rate is explained by year. e)
Do not interpret. B You must have “tends to”, “on average”, “expected”, “predicted” or other similar word. The slope is the average change in y for a one unit change in x. 4. In 2014, the poverty rate was 3.66. Find the residual. a)
2.93404 b)
-2.93404 c)
0.726 d)
-0.726
e)
0.86186 C 1.) Identify the obs x and obs y in the problem. Obs x = 2014 and Obs y = 3.66 2.) Find the predicted y by plugging obs x into the equation. Yhat = 1299.109-0.6435824(2014) = 2.934 3.) Find the residual by using the residual formula : Res= obs y – pred y . Obs y = 3.66 and Pred y = 2.934, Res = 3.66-2.934 = 0.726 5. Find the value of the correlation coefficient, r. a)
0.9284 b)
-0.9284 c)
0.7428 d)
-0.7428 B Correlation can be found by taking the square root of r-squared and then determining the sign by looking at the plot. ࠵? = −√࠵?. ࠵?࠵?࠵?࠵?࠵?࠵?
= −࠵?. ࠵?࠵?࠵?࠵?
Questions 6 - 7 Below is a scatterplot of the relationship between educational level and crime rates in Florida counties. Education level is measured as the percentage of residents over age 25 who had at least a high school degree. Crime rates is the number of crimes in that county in the past year per 1000 residents. 6.
Eight houses listed for sale in Gainesville, Florida in Spring 2017 were randomly selected. The square footages of these homes are below. Find the sample standard deviation. 1369 572 1254 984 1442 1585 1700 1708 a.)
1327 b.)
389 c.)
364 d.)
323 B See JMP instructions and output on the next page.
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Then enter y as 7.
Fill in the blank. A histogram is a graphical display to show _________ . a)
the correlation between two quantitative variables. b)
the relationship between a quantitative and a categorical variable. c)
the outcomes of categorical variable. d)
the outcomes of a quantitative variable. D A histogram only shows outcomes of one quantitative variable.
Questions 10 - 11 Suppose that a coffee machine makes 8 ounce cups of coffee. The actual amount of coffee varies from cup to cup. The amount dispersed is normally distributed with a mean of 8 oz. and a standard deviation of 0.3 oz. 10.
What is the probability that the coffee machine disperses less than 7.9 ounces of coffee? a)
0.33 b)
0.37 c)
0.44 d.) 0.66 a)
0.86 B Use JMP. Help > Sample Data > Teaching Scripts > Interactive Teaching Modules > Distribution Calculator. 11.
Find the 95
th
percentile. a)
8.49 b)
8.29 c)
8.14 d) 7.71 e) 7.51 A Use JMP. Help > Sample Data > Teaching Scripts > Interactive Teaching Modules > Distribution Calculator.
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Questions 12 – 14
Do you get better gas mileage with clean air filters? Five cars were driven with a clean air filter and with a dirty air filter. The data is below. (d= clean – dirty) Car 1 2 3 4 5 With Clean Air Filter 31 32 32 19 22 With Dirty Air Filter 30 29 29 19 20 12. What would be the alternative hypothesis? a.) Ha: ࠵?
"
<0 b.) Ha: μ
d
<0 c.) Ha: μ
d
>0 d.) Ha: ࠵?
"
>0 C If you taking clean – dirty, you would get a positive value if a clean filter resulted in higher gas mileage. Since you have the same cars that were driven with and without air filters, this data is paired. This is an example of means of dependent samples, so you would be making estimates of mu_d.
13.
Find the test statistic for this significance test. ___
3.09
____ 3.09 Enter the data into a data table in JMP. à
à
14.
What value of t would you use for a 95% confidence interval for μ
d
? _______ 2.776 Use JMP. Help > Sample Data > Teaching Scripts > Interactive Teaching Modules > Distribution Calculator. DF = n – 1 = 5-1=4
Questions 15 -16
A random sample of 40 used Toyota Priuses for sale in Atlanta, Georgia and 40 used Toyota Priuses for sale in Bangor, Maine in Spring of 2017 was taken. Assume that the cars had relatively similar mileage and were in similar condition. Assume equal variances. Variable location N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum price Atlanta 40 0 15744 311 1966 13999 14590 14981 15999 21950 Bangor, ME 40 0 15236 426 2696 11352 12616 14941 17375 20977 15. Find the standard error for the test statistic for Ho:μ
a
-μ
b = 0 versus Ha: .
Ha:μ
a
-μ
b ≠ 0.
___________ . 527.5787 Use JMP. Help > Sample Data > Calculators > Hypothesis Test for Two Means > Summary Statistics. 16. Based on the p=value from JMP, what would be a correct conclusion? a.) strong
statistically significant evidence that the population
mean used car price for Toyota Prius in Atlanta, Georgia is higher
than in Bangor, Maine. b.) no
statistically significant evidence that the population
mean used car price for Toyota Priuses in Atlanta, Georgia is different
than in Bangor, Maine. c.) strong
statistically significant evidence that the sample
mean used car price for Toyota Prius in Atlanta, Georgia is higher
than in Bangor, Maine. d.) strong
statistically significant evidence that the sample
mean used car price for Toyota Prius in Atlanta, Georgia is different
than in Bangor, Maine. e.) no
statistically significant evidence that the sample
mean used car price for Toyota Priuses in Atlanta, Georgia is different
than in Bangor, Maine.
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B Based on the p-value given in the previous problem, you would determine that you had no statistically significant evidence for HA. If no alpha is given, use the following graduated scale. p-value Range Decision p-value > 0.10 No statistically significant evidence for Ha p-value is between 0.05 and 0.10 Some statistically significant evidence for Ha p-value is between 0.01 and 0.05 Strong statistically significant evidence for Ha p-value is less than 0.01 Very Strong Statistically significant evidence for Ha 17. A study was conducted to determine if the average amount of money spent on pizza per week by students is different from 20 dollars. A random sample of 14 college students was selected. The test statistic was -1.50. Use JMP to determine the p-value. ___________ 0.1575 Use JMP. Help > Sample Data > Teaching Scripts > Interactive Teaching Modules > Distribution Calculator. For this question, you have the test statistic given, so you would go to the distribution calculator to get the p-value. Since this question is about quantitative data and you are inferring about one mean, you would use the t distribution. (Note: You couldn't use the significance test option for this problem, since you don't have the sample mean or the sample standard deviation.) 18. What is the primary purpose of the 95% confidence interval for μ
2
-μ
1 ? a.) To estimate the confidence level b.) To specify the range of values for μ
1 and μ
2 c.) To determine the difference between μ
1
-μ
2 and the null hypothesis value of 0 d.) To estimate the difference between the mean of population 1 and population 2 e.) To estimate the mean of population 1 and the mean of population 2
D – The Confidence Interval for two independent means is interested in estimating the difference between the parameters. 19. In significance testing, what does α represent? a.) The probability that the null hypothesis is true. b.) The probability that alternative hypothesis is true. c.) The probability of a Type I error d.) The probability of a Type II error e.) The probability of a measurement error C – This is the definition Questions 20 - 24
Determine what type of statistical test should be used to answer each of the questions below. (You will only use each response one time.) a.) One Mean b.) Comparing means of dependent samples c.) Comparing two independent means d.) Comparing two independent proportions e.) One Proportion 20. Are older people more likely than younger people to consider purchasing a hybrid vehicle? D – categorical Data – two groups
21. What is the typical price of a used Toyota Prius manufactured in 2011? A – quantitative data – one group 22. Is there a difference in used car price for a Toyota Prius with 2 or 4 cylinders? C – quantitative data – two groups 23. Is there a significant difference in gas mileage for a Toyota Prius driven 100 miles in the mountains versus along a flat coastline? Ten drivers determine their gas mileage for the same stretch of roads in the mountains and along the flat coastline. B – quantitative data – paired responses 24. What proportion of people purchase a hybrid cars after they have been given a test drive? E – categorical data – one group Questions 25 – 28 The General Social Survey asked participants to state if they felt that helping others was the most important thing for a child to learn to prepare him or her for life. In 2012,
out of 1304 respondents (group 1), 214 said that it was the most important. In 2016, out of 1912 respondents (group 2), 357 said that it was the most important. (dif = group 1 – group 2) 25. What is p
1
? a.) 0.186 b.) 0.164 c.) 0.178 d.) Unknown D population proportion for group 1 is unknown. The sample proportion for 2012 is given in the problem, ࠵?
࠵?
$
= 214/1304 = 0.164 26. What is(are) the symbols of the parameter that should be estimated here? a.) μ b.) p c.) p
1
-p
2
d.) ࠵?̂
"
− ࠵?̂
#
e.) ࠵?̅
"
-
࠵?̅
#
C We are estimating p
1
-p
2
by using ࠵?
*
࠵?
− ࠵?
*
࠵?
.
27. The 95% confidence interval for this parameter is (-0.049, 0.004). Finish the following sentence. “We are 95% confidence that the population proportion of Americans in 2012 who felt that helping others was the most important thing to teach a child to prepare him or her for life is . . .” a.) between 0.004 and 0.049. b.) is 0.049. c.) is between 0.049 less to 0.004 more than in 2016. d.) is between 0.049 more to 0.004 less than in 2016 e) is between 0.004 less to 0.049 less than in 2016 C Group 1 is 2012 and Group 2 is 2016. The negative sign translates into less and the positive sign translates into more. 28. Besides random sampling and categorical data, what other assumption needs to be met? a.) The number of people that said helping others was most important and not most important in 2012 and in 2016 must be at least 30. b.) The number of people that said helping others was most important and not most important in 2012 and in 2016 must be at least 10. c.) The number of people that changed their minds must be at least 30. d.) The data must come from a Normal Distribution e.) Both B and D. B We need to have at least 10 expected successes and failures in each group.
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Questions 29 -31 In the past the average life span for a flashlight battery has been 10 hours. However, the production line has been improved in the past six months. The manufacturer wants to determine if the lifetime of the battery has improved. The manufacturer tested the lifetimes of 25 batteries. 29. What would be the appropriate hypothesis to test this claim? a.) Ho: μ =0 versus Ha: μ >0 b.) Ho: μ =10 versus Ha: μ >10 c.) Ho: ࠵?̅
=0 versus Ha: ࠵?̅
>0 d.) Ho: ࠵?̅
=10 versus Ha: ࠵?̅
>10 B – Hypothesis statements are always about the population parameter. In this case, the population parameter is the population mean, µ. The researchers wanted to know if the battery life improved, so this would mean that the population mean lifetimes was higher than 10. 30. The test statistic was 2.3. What would be the p-value? ___________ 0.0152 Use JMP. Help > Sample Data > Teaching Scripts > Interactive Teaching Modules > Distribution Calculator. (see on next page) 31. What is the parameter being estimated?
a.) μ = population mean lifetime of the battery b.) ࠵?̅
= population mean lifetime of the battery c.) ࠵?̅
= sample mean lifetime of the battery d.) µ
= sample mean lifetime of the battery A – The researcher is estimating the population mean (μ). Question 32 - 34 For the hypothesis Ho: μ
1
-μ
2
=0 versus Ha: μ
1
-μ
2
≠0, the p-value equals 0.021. Let α = 0.05. 32. What decision would be made here? a.) Reject Ho b.) Fail to Reject Ho c.) Reject Ha d.) Fail to Reject Ha A Rule: p-value ≤ α Reject Ho The p-value = 0.021 < α = 0.05. 33. If an error is made here, what type of error would it be? a.) Type I error b.) Type II error c.) Type I and II error d.) No error A Since we rejected Ho, the only type of error that we could make if we made an error would be Type I error. The definition of a type 1 error is the error that occurs when we reject Ho when Ho is true. 34. Suppose that you wanted to compare these results to a confidence interval. What level confidence interval would you need to use? What value would you look for in the confidence interval? a.) confidence level = 99%, value = 0.021 b.) confidence level = 95%, value = 0.021 c.) confidence level = 99%, value = 0 d.) confidence level = 95%, value = 0 D The confidence level that goes with α = 0.05 is 95%. (1-α)100% = (1-0.05)100% = 95% You would look to see if you had similar results by looking for Zero in the confidence interval. 35. In 2010, men reported having 3.85 hours per day to relax. Is there evidence to show that in 2014, the average time that men could spend relaxing is different from 3.85? In 2014, the General Social Survey asked 602 men how many hours they spent relaxing in the past 24 hours. The 90% confidence interval was (3.591, 3.929). What can we say about the p-value? a.) It is less than 0.10. b.) It is equal to 0.10 c.) It is greater than 0.10.
d.) It cannot be determined. C – Since 3.85 is in the confidence interval, we can say that it is a plausible value for the population mean. So, this means that we should not reject Ho. We fail to reject Ho when we have large p-values. We need to have a p-value larger than alpha (α). The alpha associated with a 90% CI is 0.10. (1-α)100% = (1-0.10)100% = 90% Questions 36 -39
GenForward (a group that is a part of the survey team for the General Social Survey that concentrates on Millennials) surveyed 1836 young adults between the ages of 18-34 between June 23
rd
, 2017 and July 10, 2017 on issues related to education. One of their areas of focus was charter schools. Charter schools are publicly funded, but are privately managed. Out of the 251 Asian Americans interviewed during this time period, 153 supported charter schools. Let p equal the population proportion of Asian Americans between 18-34 years of age that support charter schools. Consider the sample of Asian American Millennials to be random and representative. 36. What is the value of p? a.) 0.64 b.) 0.61 c.) 0.14 d.) 0.08 e.) Unknown E population proportion is unknown. The sample proportion is given in the problem, ࠵?
*
= 153/251 = 0.61 37. What would be the margin of error for a 95% confidence interval for p? a.) 0.01 b.) 0.02 c.) 0.03 d.) 0.06 e.) Unknown D ࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?࠵?࠵?࠵?࠵? = ࠵? 7
࠵?
"(࠵?%࠵?
")
࠵?
= ࠵?. ࠵?࠵?7
࠵?.࠵?࠵?࠵?࠵?(࠵?%.࠵?࠵?࠵?࠵?)
࠵?࠵?࠵?
= 0.06 38. What would be the value of z for a 91.5% confidence interval for p? a.) 1.37 b.) 1.72 c.) 1.79 d.) 3.47 e.) -35.7 B Use JMP. Help > Sample Data > Teaching Scripts > Interactive Teaching Modules > Distribution Calculator.
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39. Suppose that a new survey about support for charter schools was to be conducted this December also of randomly selected Millennial Asian Americans. This time they wanted to have a margin of error of 0.10, at 95% confidence. They wanted to use information from this summer as an estimate of the population proportion. What is the desired sample size? ______________ 92 Use JMP. Help > Sample Data > Calculators > Sample Size for Confidence Intervals > Proportions. Questions 40 - 41
Michelin Defender Tires have an average lifetime of 90000 miles with a standard deviation of 5000 miles. Suppose that 60 tires were randomly selected (from different vehicles and positions on the car). 40. What would be the mean and standard deviation of the sampling distribution of the sample mean lifetime of the tires? a.) mean = 90000 and standard deviation = 5000 b.) mean = 90000 and standard deviation = 645.5 c.) mean = 90000 and standard deviation = 125 d.) mean = unknown and standard deviation = 645.5 e.) mean = unknown and standard deviation = unknown B The Central Limit Theorem states that the sampling distribution will be normally distributed with a mean of µ and a standard deviation of ࠵?
√࠵?
. In this problem µ = 90000 and σ = 5000. So, the standard deviation is ࠵?࠵?࠵?࠵?
√࠵?࠵?
= 645.5 41. If the sample size was increased, what would happen to the mean of the sampling distribution of the sample mean? a.) It would stay the same. b.) It would increase. c.) It would decrease. d.) Cannot be determined. A As the sample size increases, the mean of the sampling distribution of the sample mean stays the same. It is always equal to µ, the population mean.
Questions 42 – 43 Also, in the Genforward August Report about education was a survey that asked if parents favored or opposed allowing public school parents to excuse their children from taking standardized state tests. Out of the 506 African Americans between the ages of 18 and 34 surveyed, 283 opposed this. Consider this sample to be random and representative. Is this significance evidence to show that the population proportion of African Americans between the ages of 18-34 who oppose is more than 0.50? 42. What is the appropriate alternative hypothesis? a.) Ha: p = 0.50 b.) Ha: p > 0.50 c.) Ha: p > 0.56 d.) Ha: ࠵?̂
= 0.50 e.) Ha: ࠵?̂
> 0.50 B – The researcher wants to know if the population proportion is greater than 0.50. You never use sample statistics or the value of the sample statistics in hypothesis statements. 43. What would be the test statistic? _____________ TS = 2.6673 Use JMP. Help > Sample Data > Calculators > Hypothesis Test for One Proportion > Summary Statistics. 44. If everything else remains the same, what happens as the confidence level is increased for a confidence interval for the population proportion? a.) The width of the interval increases. b.) The width of the interval decreases. c.) The width remains the same. d.) Not possible to determine. A As the confidence level increases, the width of the confidence interval increases. This way we are more confident that we capture the population parameter.
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45.
A new type of flu vaccine is being developed that manufactures the vaccine in a cell culture rather than in chicken eggs. A clinical trial of 3,900 randomly selected participants was performed. Each participant was given the new type of flu vaccine. Out of the 3,900 participants given this vaccine, 26 had the flu in the following 28 weeks. Is this evidence that the population proportion that would get the flu after receiving this vaccine is different from 0.01? The test statistic is -2.09. What is the p-value? ____________ 0.0366 Use JMP. Help > Sample Data > Teaching Scripts > Interactive Teaching Modules > Distribution Calculator. For this question, you have the test statistic given, so you would go to the distribution calculator to get the p-value. Since this question is about categorical data and you are inferring about one proportion, you would use the z distribution. (Note: For this problem, you could put the data into the significance test for one proportion, since you are given all the information.)
46. For a significance test for the population proportion with significance level equal to 0.10 and p-value equal to 0.06, what would be the correct decision? a.) Reject Ho b.) Fail to Reject Ho c.) Reject Ha d.) Fail to Reject Ha A Rule: p-value ≤ α Reject Ho – Since the p-value is less than alpha, we would reject Ho. If the p-value was larger than alpha, we would have failed to reject Ho. C and D are never appropriate responses. 47. Suppose that the probability that a baby girl is born is 0.50. Suppose that there were two hospitals. Hospital A had about 200 births a month and hospital B had about 50 births a month. Assuming that these samples could be considered random samples, which is more likely to have a sample proportion of girls higher than 0.53? a.) Hospital A b.) Hospital B c.) It would be the same. B – Hospital B has a higher probability of the sample proportion being larger than 0.53. Hospital A Sampling Distribution of the sample proportion is approx.. Normal with mean = 0.50 and standard deviation = 7
࠵?.࠵?(࠵?%࠵?.࠵?
࠵?࠵?࠵?
= ࠵?. ࠵?࠵?࠵?࠵?
Use JMP. Help > Sample Data > Teaching Scripts > Interactive Teaching Modules > Distribution Calculator. The probability equals 0.1977. (picture below). Hospital B Sampling Distribution of the sample proportion is approx.. Normal with mean = 0.50 and standard deviation = 7
࠵?.࠵?(࠵?%࠵?.࠵?
࠵?࠵?
= ࠵?. ࠵?࠵?࠵?࠵?
Use JMP. Help > Sample Data > Teaching Scripts > Interactive Teaching Modules > Distribution Calculator. The probability equals 0.3357. (picture below). Hospital A
Hospital B 48. Is the below statement true or false? “The p-value is the probability that the null hypothesis is true.” a.) True b.) False B False – The definition of the p-value is the probability that we see a test statistic this extreme or more if the null hypothesis is true. The statement above is not complete. 49. Which of the following is not
correct? The standard deviation of a statistic describes a.) The variability in the values of the statistic for repeated samples of size n b.) How close the statistic falls to the parameter that it estimates c.) The standard deviation of the sample data measurements
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d.) The standard deviation of the sampling distribution of that statistic C – The standard deviation of a statistics refers to how much the statistics changes. Answer C is talking about the sample data measurements not the statistics. The difference in the answers is observation units under investigation. This question is asking about the standard deviation of the statistic
In answers a, b, and d, the observational units are the statistics
. a.) The variability in the values of the statistic
for repeated samples of size n b.) How close the statistic
falls to the parameter that it estimates d.) The standard deviation of the sampling distribution of that statistic
So, a, b, and d are all correct. However, in answer c, the observational units are the raw data values. c.) The standard deviation of the sample data measurements
The observational units let you know which is correct or incorrect. Questions 50 - 51
In the GenForward survey, there was also a question that asked Caucasian Millennials if they agreed with the following statement. “College is necessary to be successful in today’s work world”. Assume that the sample was random and representative. The 95% confidence interval for p was (0.3961, 0.4823). Please mark each of the interpretations as either A for Correct interpretation or B for Incorrect interpretation. 50. The probability that p is between 0.3961 and 0.4823 is 0.95. a.) This is a correct interpretation. b.) This is an incorrect interpretation. B Incorrect. You can’t state specific values of the confidence interval and use specific values in the confidence interval. You can only speak of probability in the general sense. For example, “The probability that the true proportion is in a 95% confidence interval is 0.95” is correct. 51. We are 95% confident that the true proportion of Caucasian Millennials who agree that college is necessary to be successful in today’s work world is between 0.3961 and 0.4823. a.) This is a correct interpretation. b.) This is an incorrect interpretation. A This is correct. It talks about estimating the population proportion and uses the term confidence, not probability.
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Questions 52 – 54 In 1998, a scientist did a chemical analysis of a random sample of wines grown in the same region in Italy. For the 178 different wines in the sample, the mean of fixed acidity was 13.00 g/dm
3
with a standard deviation of 0.811 g/dm
3 . 52. What parameter could be estimated with this data? a.) ࠵?̅
= the sample mean fixed acidity of the 178 samples b.) ࠵?̅
= the population mean fixed acidity of the wines from this region c.) ࠵?
= the sample mean fixed acidity of the 178 samples d.) ࠵?
= the population mean fixed acidity of the wines from this region D The parameter is µ, the population mean. In confidence intervals we try to estimate the parameter. The sample statistic, the sample mean(
࠵?
A)
, is known we don’t estimate it. 53. Compute the 95% confidence interval for the parameter. What is the upper limit of the 95% confidence interval? ____________________ 13.12 Use JMP. Help > Sample Data > Calculators > Confidence Interval for the Population Mean > Summary Statistics. 54. Compute the 95% confidence interval for the parameter. What is the lower limit of the 95% confidence interval? ____________________ 12.88 Use JMP. Help > Sample Data > Calculators > Confidence Interval for the Population Mean > Summary Statistics. (see problem above).
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55. Suppose that they wanted to redo this analysis in 2018. They wanted to have a margin of error of 0.10 and use this study to estimate the value of the standard deviation. They want to be 95% confident. What sample size would they need? ____________________ 253 Use JMP. Help > Sample Data > Calculators > Sample Size for Confidence Intervals > Mean. 56. Which of the following is not
a characteristic of the t distribution? a.) symmetric b.) continuous c.) centered at zero d.) has more spread in the tails for smaller degrees of freedom e.) All of the above are correct E – These are all characteristics of the t distribution. 57. Do more than 80% of Hispanics feel that tuition to public colleges should be free? In the GenForward report, 521 Hispanic Millennials (consider this a random and representative sample) were asked if tuition to public colleges should be free. Out of the 521, 443 said that it should be free. The test statistic equals 2.87. What is the p-value? _______________ 0.0021 Use JMP. Help > Sample Data > Teaching Scripts > Interactive Teaching Modules > Distribution Calculator.
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OR Use JMP. Help > Sample Data > Calculators > Hypothesis Test for One Proportion 58. For the sampling distribution of the sample mean, what happens to the standard deviation as the sample size increases? a.) It increases. b.) It decreases. c.) It would stay the same. d.) It would vary based on the distribution. B It gets smaller. The formula is 0
√1
.
As n increases, the quantity decreases. 59. For a 98% confidence interval for the population mean, what value of t would you use? The sample size is 17. ________________ 2.583 Use JMP. Help > Sample Data > Teaching Scripts > Interactive Teaching Modules > Distribution Calculator. Df = n – 1 = 17-1 = 16
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Questions 60-62
Only 1.5% of the human population has red hair. Suppose that a random sample of 2000 people from across the world is selected. In this sample of 2000 people, 34 people had red hair. 60. What would be the mean of the sampling distribution of the sample proportion? a.) 0.015 b.) 0.017 c.) 0.15 d.) 0.17 e.)
Unknown A – The mean of the sampling distribution of the sample proportion is p. The true proportion (p) is 0.015. 61. What would be the approximate distribution of the sampling distribution of the sample proportion? a)
Binomial b)
Normal c)
Skew Left d)
Unknown B Since np and n(1-p) are both greater than 10, we would say that the sampling distribution of the sample proportion is Normally distributed. (n =2000 , p 0.015 so np = 2000*0.015 = 30 and n(1-p) = 2000(1-0.015) = 1970. 62. What would be the standard error of the sampling distribution of the sample proportion? ________ 0.0027 The standard error of the sampling distribution of the sample proportion is 7
࠵?࠵?
࠵?
= 7
࠵?.࠵?࠵?࠵?࠵?(࠵?%࠵?.࠵?࠵?࠵?࠵?)
࠵?࠵?࠵?࠵?
= ࠵?. ࠵?࠵?࠵?࠵?
63 -65 To be competitive in global markets, many US corporations are undertaking major reorganizations. Often these involve “downsizing” or a “reduction in force” (RIF), where substantial
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numbers of employees are terminated. Federal and various state laws require that employees be treated equally regardless of their age. In particular, employees over the age of 40 years are in a “protected” class, and many allegations of discrimination focus on comparing employees over 40 with their younger coworkers. Here are the data for a recent RIF: Standing Over 40 Under 40 Total Terminated 16 78 94 Non Terminated 585 765 1350 Total 601 843 1444 63. What is the expected count for those that are over 40 and terminated under the null hypothesis? ______ 39.123 To find the expect counted count, you first need to find the row and column totals. Expected Count = ࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?∗࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?
࠵?࠵?࠵?࠵?࠵?
=
࠵?࠵?∗࠵?࠵?࠵?
࠵?࠵?࠵?࠵?
= ࠵?࠵?. ࠵?࠵?࠵?
64. What is the contribution to the test statistic of the under 40/ not terminated category? a)
0.6784 b)
-0.6784 c)
0.0293 d)
-0.0293 A First find the observed and expected counts. Observed = 765 Expected Count = ࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?∗࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?
࠵?࠵?࠵?࠵?࠵?
=
࠵?࠵?࠵?࠵?∗࠵?࠵?࠵?
࠵?࠵?࠵?࠵?
= ࠵?࠵?࠵?. ࠵?࠵?࠵?
(࠵?࠵?࠵? − ࠵?࠵?࠵?)
࠵?
࠵?࠵?࠵?
=
(࠵?࠵?࠵? − ࠵?࠵?࠵?. ࠵?࠵?࠵?)
࠵?
࠵?࠵?࠵?. ࠵?࠵?࠵?
= ࠵?. ࠵?࠵?࠵?࠵?
65. What is the sampling distribution for this problem? a)
Chi Square with 1 degree of freedom b)
Chi Square with 2 degrees of freedom c)
Chi Square with 3 degrees of freedom d)
t with 1 degree of freedom e)
t with 3 degrees of freedom A This is a Chi Square Test so, the df = (r-1)(c-1). So, the df = (2-1)(2-1) = 1* 1 = 1 Questions 66-67
“Professor Green never uses C”. One of the employees of an off-campus tutoring service claims that Management professor Richard Green sets his exams up so that a “C” response is the correct answer only about once on each 25-question exam. The other four responses (“A”, “B”, “D” and “E”) all show up with equal frequency. Professor Green hears about this claim but does not believe he
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does that. He looks at a number of exams he has given over the past few years and tallies the correct answers. The results are in the table at the right. 66.The tutor-service employee claimed one answer in 25 was “C” and the rest occurred equally-
often. What is the expected count for C under this assumption? One in 25 = 1/25 = .04 or 12 per 300 The expected count would be 12. 67. See the above completed table, it shows the work necessary to find the Chi Square test statistic which equals 8.2778. But, which of the five possible exam response choices was relatively most different from what was expected under the tutor-service employee’s claim? A. choice A B. choice B C. choice C D. choice D E. choice E Answer Choice A: Since the contribution to the Chi Square Test is the largest. Questions 68 - 70
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68.
Use the data set that appears on the previous page. Assume you want to compare the average CARS score for Finance
majors to those of Marketing
majors. You may assume the variances in scores for these two majors are about equal. Under this assumption, what is the test statistic? ________ (Finance – Marketing) -2.6791 Use JMP. Help > Sample Data > Calculators > Hypothesis Test for Two Means > Summary Statistics. Since they take "Finance - Marketing", this would make Finance group 2 and marketing group 1. Finance: xbar2 = 55.349, s2 = 9.897 and n2 = 43 Marketing: xbar1= 61.50 s1 = 8.021 n1 = 26 TS = -2.6791.
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69.
Again assume you are comparing the average CARS score for Finance
majors to those of Marketing
majors. If the significance level (α) = 0.05, what decision do you make? a.) Reject Ho b.) Fail to Reject Ho c.) Reject Ha d.) Fail to Reject Ha A Rule: p-value ≤ α Reject Ho – Since the p-value is less than alpha, we would reject Ho. If the p-
value were larger than alpha, we would have failed to reject Ho. C and D are never appropriate responses. 70.
A CARS scores of 65 or above may show a relatively high degree of anxiety. Suppose we wish to estimate the proportion of all Finance
majors that have this level of computer anxiety. If you make a 95% interval estimate for this proportion, what is the interval’s margin of error (
ME
)? 0.096 The margin of error is everything after the ± in the CI. Look at the data set on the previous page and count the number of values that are equal to or higher than 65. There are five values (65, 66, 66, 67, 89). We will then have X = 5 and n = 43. Phat = sample proportion = 5/43 = 0.11628 Margin Error = 1.96*sqrt(.11628(.88372)/43)= 1.96*.0488=.096
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Questions 71- 73 If the airline flight that you are on is 20 minutes late departing, can you expect the pilot to make these minutes up by, say, flying faster than usual? Data on 40 flights out of Atlanta heading west was analyzed. Recorded were the delay in departure (in minutes) and the delay in arrival (also in minutes). A few flights had early departures or arrivals, and these are recorded as a negative delay. For example, a departure delay of -5 means the flight departed 5 minutes early. The data are in the file AirlineDelays.xlsx. Run a simple regression predicting arrival delay from departure delay. 71.
What is R
2
? __
0.927878
________
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72.
Even though the data is linearly related, find Spearman’s rho. ____________ Spearman’s rho = 0.8964 73. Look at the residuals and determine which of the following statements correctly discusses the condition of equal spread? a)
The errors are nearly normally distributed, so the equal spread condition is met. b)
The errors have an extremely skewed distributed, so the equal spread condition is not met. c)
The residual plots clearly display a funnel shape, so the equal spread condition is not met. d)
The residual plots do not display a funnel shape, so the equal spread condition is met.
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No funnel shape, so answer is D. 74. Why might you use Spearman’s rho? a.) Because the data is not linearly related, but you want to discuss the association. b.) Because the data is categorical, but you want to discuss the association. c.) Because the data is not randomly selected, but you want to discuss the association. d.) Because the residuals don’t have a nearly normal distribution, because you want to discuss the association. A – Spearman’s rho does not require that the x and y variables have a linear relationship.
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Questions 75- 76
Twelve runners are asked to run a 10-kilometer race on each of two consecutive weeks. The same runners wear one brand of shoe in one race and in the other race a second brand. Which brand they wear in which race is determined at random. All runners are timed and are asked to do their best in each race. The Minitab results are below. Paired T for Brand 1 - Brand 2 95% CI for mean difference: (-0.441, 0.079) 75. Give the best interpretation of the confidence interval. a.)
We are 95% confident that when the runners ran with brand 1 sneakers, they ran at least 0.079 minutes to at most 0.441 minutes faster than when the runners ran with brand 2 sneakers, on average. b.)
We are 95% confident that when the runners ran with brand 2 sneakers, they ran at least 0.079 minutes to at most 0.441 minutes faster than when the runners ran with brand 1 sneakers, on average. c.)
We are not confident that there is a difference between the average runs’ times when brand 1 and brand 2 were worn. d.)
We are 95% confident that the run times for brand 1 were between –0.441 and 0.079. C– You want to make sure that you pay attention to the order of subtraction, so the comparison is from brand 1 to brand 2. 76. State the null and alternative hypothesis to test to see if brand 1 is faster than brand 2. a.) Ho: μ
d
=0 Ha: μ
d <0 b.) Ho: μ
d
=0 Ha: μ
d >0 c.) Ho: μ
1
-μ
2
=0 Ha: μ
1
-μ
2
>0 d.) Ho: μ
1
-μ
2
=0 Ha: μ
1
-μ
2
<0 A -Since this is paired data, you would use μ
d . Since the data is brand 1 – brand 2, if they believe that brand 1 results in a faster pace than brand 2, this belief would result in the time for brand 1 being lower than time for brand 2. This would indicate that brand 1 – brand 2 < 0. So, the correct answer is a. 77.
Which of the following is a hypothesis statement for paired means? a.)
Ho: µ= 0, Ha: µ ≠ 0 b.)
Ho: p = 0, Ha: p ≠ 0 c.)
Ho: µ
D
= 0, Ha: µ
D
≠ 0 d.)
Ho: µ
1
= µ
2
, Ha: µ
1
≠ µ
2
C
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Questions 78 – 81 Determine what type of statistical test should be used to answer each of the questions below. (You will only use each response one time.) a.) Chi Square Goodness of Fit Test b.) Chi Square Test of Homogeneity c.) Chi Square Test of Independence d.) Simple Linear Regression 78.A local café owner wanted to predict the sales of pizza throughout the Spring semester. He wanted to accurately estimate how much pizza would be purchased each day. D This problem has a quantitative x variable (time) and quantitative y variable (sales of pizza). When you have two quantitative variables and you are interested in prediction, this is an example of simple linear regression. 79. The same local café owner wanted to know if there was a difference in the number of coffee purchases between 7 and 10am each morning. He had initially assumed that each day of the week (Sunday – Saturday) would have the same number of purchases. Is there evidence that the number of coffee purchases has a different pattern than expected? A There is only one sample. The café owner is looking to see if the observed pattern is different from expected. This is a Goodness of Fit question. 80.
The local café owner gave out a two question survey that asked customers if they would visit the café if they were open between 5 - 6am, 6 – 7am, both or neither. He also asked if they would visit the café’ if they were open between 8 - 9pm, 9 – 10pm, both or neither. C This has only one sample; however, there are two questions. the cafe owner wants to know if the responses are different for the two questions. So, this is a Chi Square test of Independence. 81.
The local café owner sent out a one question survey to his preferred customers that were high school students, college students, and those that weren’t students. He asked them if they would prefer to get a free cup of coffee after 10 visits, or a free cup of coffee and a doughnut after 15 visits. Is there a difference in responses? B This question has three groups: high school students, college students and non students. He wants to know if there is a difference in the responses to one question. So, this is Chi Square Test of Homogeneity. Summary of the Three Types of Chi Square Distributions The Chi Square Goodness of Fit test is when you have only one sample of results with one response for case; however the responses come from a possible set of multiple categorical responses. You are looking to see if the proportion in each group is different from hypothesized proportions. For example, do 1/3rd of UF students prefer white, milk and dark chocolate? The Chi Square Test of Homogeneity has two or more groups with only one response from each person, but these responses can be any one of multiple
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categories. Is there a difference in chocolate preferences (white, dark or milk) between UF and FSU undergraduates? The Chi Square Test of Independence has two responses from each individual. The responses come from multiple categories. For example, UF students are asked if they strongly agree, agree, neutral, disagree, or strongly disagree with two political statements. Are those responses independent? Questions 81 – 85 In 2016, the General Social Survey included a question that asked males and females if they were to get enough money to live as comfortably as you would like for the rest of your life, would you continue to work or would you stop working? These questions were asked to a random sample of 1,039 adults living in the US. Is there a significant difference in how males and females responded to this question. Continue Working Stop Working Total Male 371 134 505 Female 376 158 534 Total 747 292 1039 82.
What type of tests would this be? a.
Two Independent Means b.
Chi Square Test of Homogeneity c.
Simple Linear Regression d.
One means e.
One proportion B Since this is categorical data comparing at least two groups, this is Chi Square Test of Homogeneity. 83.
Find the expected count for men who would continue working. ____
363.075
____________ To run this in JMP, enter the data into JMP by creating three columns: Gender, Response, and Frequency. Then, go to Fit Y by X. Enter Gender as X, Response as Y and Freq as Frequency. On the output, click on the red triangle next to contingency table and select expected and chi square contribution.
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84.
Find the contribution to the Chi Squared test statistic for men who continue working. ___
0.1730
__________ 85.
Which cell has the highest contribution to the Chi-Square Test statistic? a.)
Male/Continue Working b.)
Female / Continue Working c.)
Male / Stop Working d.)
Female / Stop Working C The highest contribution to the Chi Square Test Statistic is 0.4425.
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86.
Based on the p-value what conclusion would you make? B The p-value is 0.2738. The p-value is larger than alpha, so we don’t have evidence that the responses are different. a.)
With a p-value higher
than 0.05, we have evidence that there is a difference in how males and females in the US respond to this question. b.)
With a p-value higher
than 0.05, we don’t have evidence that males and females in the US answered this question differently. c.)
With a p-value lower
than 0.05, we have evidence that there is a difference in how males and females in the US respond to this question. d.)
With a p-value lower
than 0.05, we don’t have evidence that males and females in the US answered this question differently. 87.
Besides random sampling, what other assumption needs to be met? a.)
The population distribution is normally distributed. b.)
There are at least 5 expected in each cell. c.)
The sample distribution is normally distributed. d.)
A and B e.)
B and C B For Chi square test, you need to have the expected cell count to be at least 5. The data is categorical, so the data will never exhibit a normal distribution. 88.
What other testing method would have given equivalent results? a.)
One Proportion Test with a two sided alternative hypothesis b.)
One Proportion Test with a one sided alternative hypothesis c.)
Two Independent Proportion Test with a two sided alternative hypothesis d.)
Two Independent Proportion Test with a one sided alternative hypothesis C The Chi Square Test for Homogeneity between two groups is the same as the Two Independent Proportion Test for a two sided hypothesis – to see if the population proportions are different between the two groups. 89. In a study of 1991 model cars, a researcher computed the least-squares regression line of price (in dollars) on horsepower. He obtained the following equation for this line. yhat = –6677 + 175 × horsepower Based on the least-squares regression line, what would we predict the cost of a 1991 model car with horsepower equal to 200 to be? yhat = –6677 + 175 × 200 = 28323 dollars
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Questions 88 – 89
90. Using the data set, GainesvilleHomes_Sp2019, find the least squared regression line to predict price(thousands) based on square footage. In JMP. Fit Y by X. Select Square footage as X and price(thousands) as y. The regression equation is yhat = -152.7785 + 0.2206449x 91.
Using the above regression line, what is the proportion of variation of price(thousands) that is explained by linear regression on square footage? R-squared = 0.857 92.
In a Chi Square Goodness of Fit test, if there is a large difference between the observed and expected counts, what happens to the test statistic and the p-value? a.)
Large test statistic, large p-value b.)
Large test statistic, small p-value c.)
Test statistic close to 1, large p-value d.)
Test statistic close to 1, small p-value B The top of the test statistics is the sum of the difference between the observed and the expected for each cell, squared. If there is a large difference between the observed and the expected counts, the test statistic becomes larger. Additionally, since you always shade greater than, this results in the p-
value becoming smaller.
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Questions 93-96 Is there a difference in the population mean number of text messages sent for male and female students at UF? Students at UF were randomly given a survey asking about their texting habits. The data is below. (Difference: Male - Female) Mean Standard Deviation Sample Size Females 95.699 143.61 163 Males 61.63 65.31 57 93.
What is the point estimate of difference between the population mean number of texts between male and female students at UF? a.)
-34.069 b.)
34.069 c.)
-2.76 d.)
2.76 e.)
Unknown value A 61.63-95.699= -34.069 94.
What would be the correct alternative hypothesis? a.)
࠵?࠵?: ࠵?̅
C
− ࠵?̅
D
≠ 0
b.)
࠵?࠵?: ࠵?̅
C
− ࠵?̅
D
> 0
c.)
࠵?࠵?: ࠵?
C
− ࠵?
D
> 0
d.)
࠵?࠵?: µ
C
− µ
D
≠ 0
e.)
࠵?࠵?: µ
C
− µ
D
< 0
D 95.
If the p-value equals 0.0173, what statement can you make about the confidence interval for the difference in the population mean number of text messages? a.)
The 95% confidence interval would include 0. b.)
The 99% confidence interval would not include 0. c.)
The 95% confidence interval would not include 0. d.)
The 90% confidence interval would include 0. At the alpha = 0.05 level, you would reject Ho:
࠵?࠵?: µ
C
− µ
D
≠ 0
. This would indicate that a 95% confidence interval would not include zero. At the alpha = 0.01 level, you would fail to reject Ho. This would indicate that zero would be in the interval. At the alpha = 0.10 level, you would reject Ho. This would indicate that zero would not be in the interval.
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96. If the p-value is 0.0173, what can you conclude? a.)
There is strong evidence of a difference in the population mean number of text messages sent for men and women. b.)
There is strong evidence of a difference in the sample mean number of text messages sent for men and women. c.)
There is no statistically significant evidence of a difference in the population number of text messages sent for men and women. d.)
There is no statistically significant evidence of a difference in the sample mean number of text messages sent for men and women. A Since there is a small p-value, you would conclude that there was a difference in the population mean text messages. 97. Can we assume equal variances? a.) Yes, because the means are almost the same. b.) Yes, because the difference in the means is less than 2. c.) No, because the ratio of the standard deviations (s
F
/ s
m
) is greater than 2. d.) Yes, because the ratio of the standard deviations (s
F
/ s
m
) is less than 2. C 143.61/65.31 = 2.19 so we can’t assume equal variances. Questions 98 – 100 Ebony wanted to determine if the number of sales per day at her bakery were like last year. Last year, her sales on Friday, Saturday, and Sunday were twice those of the other days in the week. She is open seven days a week? 98. What type of test would she need to confirm that this year was different from last year? a.) Two Independent Proportions b.) Chi Square Goodness of Fit Test c.) Chi Square Test of Independence d.) Chi Square Test of Homogeneity B This is a Chi Square Goodness of Fit test, because she is comparing a current distribution to a historic distribution. 99. What would be the degrees of freedom for the appropriate test? ____
6
_____
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There are 7 days of the week, so there are seven categories. The degrees of freedom are equal to n-1 = 7-1=6 100. If Ebony typically had 1000 sales per week, what would be the expected number of sales on Sunday? ___
200
________ Sunday would be expected to get 2/10 of the sales. If there are 1000 sales per week, 1000* (2/10) = 200.
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