Analyzing Reaction Times and Salary Impacts of MBA on Managers

Question 1 1 / 1 point A researcher is testing reaction times between the dominant and non-dominant hand. They randomly start with each hand for 20 subjects and their reaction times in milliseconds are recorded. Test to see if the reaction time is faster for the dominant hand using a 5% level of significance. The hypotheses are: H 0 : μ D = 0 H 1 : μ D > 0 t-Test: Paired Two Sample for Means Non- Dominant Dominant Mean 63.33 56.28 Variance 218.96431 58 128.75221 05 Observations 20 20 Pearson Correlation 0.9067 Hypothesized Mean Difference 0 df 19

t Stat 4.7951 P(T<=t) one-tail 0.0001 t Critical one-tail 1.7291 P(T<=t) two-tail 0.0001 t Critical two-tail 2.0930 What is the correct decision? Reject H 0 Accept H 0 Reject H 1 Accept H 1 Do not reject H 0 Hide question 1 feedback The p-value for a one tailed test is 0.0001. This is given to you in the output. No calculations are needed. 0.0001 < .05, Reject Ho, this is significant. Question 2 1 / 1 point A manager wants to see if it is worth going back for an MBA degree. They randomly sample 18 managers' salaries before and after undertaking a MBA degree and record their salaries in thousands of dollars. Assume Salaries are normally distributed. Test the claim that the MBA degree, on average, increases a manager's salary. Use a 10% level of significance. t-Test: Paired Two Sample for Means

New Salary Old Salary Mean 59.82778 52.66667 Variance 175.5551 112.7012 Observations 18 18 Pearson Correlation 0.7464 Hypothesized Mean Difference 0 df 17 t Stat 3.4340 P(T<=t) one-tail 0.0016 t Critical one-tail 1.3334 P(T<=t) two-tail 0.0033

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

t Critical two-tail 1.7396 The hypotheses for this problem are: H 0 : μ D = 0 H 1 : μ D > 0 What is the correct p-value? 0.0033 0.0016 -1.3334 -3.4340 0.7464 Hide question 2 feedback No calculations are needed. The answer is given to you in the output. P(T<=t) one-tail 0.0016 Question 3 1 / 1 point A math teacher tells her students that eating a healthy breakfast on a test day will help their brain function and perform well on their test. During finals week, she randomly samples 45 students and asks them at the door what they ate for breakfast. She categorizes 25 students into Group 1 as those who ate a healthy breakfast that morning and 20 students into Group 2 as those who did not. After grading the final, she finds that 48% of the students in Group 1 earned an 80% or higher on the test, and 40% of the students in Group 2 earned an 80% or higher. Can it be concluded that eating a healthy breakfast improves test scores? Use a 0.05 level of significance. Hypotheses: H 0 : p 1 = p 2

H 1 : p 1 > p 2 In this scenario, what is the p-value? (Round to 4 decimal places) p-value = ___ 0.2958 ___ Hide question 3 feedback z =.48−.40.4444 ∗ .5556 ∗ (1/25+1/20) z = 0.536656 Use NORM.S.DIST(0.536656,TRUE) to find the for the lower tailed test. 0.704247, then 1 - 0.704247, this is the p-value you want to use for an upper tailed test. Question 4 1 / 1 point The manager at a sports radio station will be covering several football games over the weekend. She knows that most of her listeners are at least 22 years old and wants to know what age group she should gear her advertisements to for the games. She takes a random sample of 45 listeners from age 22-39 and 55 listeners who are 40 and older and asks them if they are likely to tune in to football games. The "yes" responses are recorded below. Age 22-39 (Group 1) Age 40+ (Group 2) Responded "Yes" to Football 32 44 Sample Size n 45 55 At the 0.05 level of significance, we are attempting to investigate if there is a significant difference in the proportion of listeners based on age. Enter the P -Value - round to 4 decimal places.

p-value = ___ 0.3005 ___ Hide question 4 feedback This is a two tailed test because you want to find significant difference. z = .71111−.80.76 ∗ .24 ∗ (1/45+1/55) z = -1.03543 Use NORM.S.DIST(-1.03543,TRUE) to find the for the lower tailed test. This value is smaller of the two, thus you will multiply it by 2 for a two tailed test 0.150233*2, this is the p-value you want to use for the conclusion. Question 5 1 / 1 point Two competing toothpaste brands both claim to produce the best toothpaste for whitening. A dentist randomly samples 48 patients that use Brand A (Group 1) and finds 30 of them are satisfied with the whitening results of the toothpaste. She then randomly samples 45 patients that use Brand B (Group 2) and finds 33 of them are satisfied with the whitening results of the toothpaste. Construct a 99% confidence interval for the difference in proportions and use it to decide if there is a significant difference in the satisfaction level of patients. Enter the confidence interval - round to 3 decimal places. ___ -0.356 ___ (50 %) < p 1 - p 2 < ___ 0.139 ___ ( 50 %)

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Hide question 5 feedback Z-Critical Value =NORM.S.INV(.995) = 2.575 LL = (.625-.7333) - 2.575*.625 ∗ .37548+.7333 ∗ .266745 UL = (.625-.7333) + 2.575*.625 ∗ .37548+.7333 ∗ .266745 Question 6 1 / 1 point Match the Symbol with the correct phrase. __ 1 __ Reduced __ 4 __ More than __ 2 __ At most __ 3 __ At least __ 5 __ Different from 1 . < 2 . ≤ 3 . ≥ 4 . > 5 . ≠ Hide question 6 feedback Reduced is less than At most means you can not have any more than that amount. That is the most you are allowed to have. At least, this is the least amount you can have but you can have more than the least amount. More than is greater Different from is not equal Question 7 1 / 1 point A null hypothesis can only be rejected at the 5% significance level if and only if: a 95% confidence interval includes the hypothesized value of the parameter a 95% confidence interval does not include the hypothesized value of the parameter

the null hypotheses includes sampling error the null hypothesis is biased Question 8 1 / 1 point A lab technician is tested for her consistency by taking multiple measurements of cholesterol levels from the same blood sample. The target accuracy in the past has been 2.5 or less with a standard deviation of 1.15. If the lab technician takes 16 measurements and the variance of the measurements in the sample is 2.2, does this provide enough evidence to reject the claim that the lab technician's accuracy is within the target accuracy? State the null and alternative hypotheses. H 0 : µ ≠ 2.5, H 1 : µ = 2.5 H 0 : µ ≤ 2.5, H 1 : µ > 2.5 H 0 : µ ≥ 2.5, H 1 : µ < 2.5 H 0 :µ < 2.5, H 1 : µ ≠ 2.5 Hide question 8 feedback The hypothesized value is 2.5 and the keyword is less which makes this a lower tailed test. Question 9 1 / 1 point Results from previous studies showed 79% of all high school seniors from a certain city plan to attend college after graduation. A random sample of 200 high school seniors from this city reveals that 162 plan to attend college. Does this indicate that the percentage has increased from that of previous studies? Test at the 5% level of significance. What is the p-value associated with your test of hypothesis? 0.7563 0.6874

0.4874 0.2437 Hide question 9 feedback Hypothesized value is .79 and this is an upper tailed test because of the keyword increased z = .81−.79.79 ∗ .21200 z = 0.694419 NORM.S.DIST(0.694419,TRUE) = 0.75629. This is the p-value for a lower tailed test. To find the p-value for an upper tailed take 1 - 0.75629. Use this value for the conclusion Question 10 1 / 1 point The alternative hypothesis is also known as the: research hypothesis null hypothesis elective hypothesis optional hypothesis Question 11 1 / 1 point A movie theater company wants to see if there is a difference in the average movie ticket sales in San Diego and Portland per week. They sample 20 sales from San Diego and 20 sales from Portland over a week. Test the claim using a 5% level of significance. Assume the variances are unequal and that movie sales are normally distributed. San Diego Portland

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

221 209 221 214 206 223 213 221 226 218 243 214 182 222 229 220 214 223 233 222 231 233 217 219 219 226

234 226 239 219 235 228 211 211 239 216 221 219 234 209 The Hypotheses for this problem are: H 0 : μ 1 = μ 2 H 1 : μ 1 μ 2 Find the p-value. Round answer to 4 decimal places. Make sure you put the 0 in front of the decimal. p-value =___ ___ Answer: 0.2853 Hide question 11 feedback Use the Data Analysis Toolpak in Excel.

Data -> Data Analysis -> scroll to where is says t:Test 2 Samples Assuming Unequal Variances -> OK Variable Input 1: is San Diego Variable Input 2: is Portland The Hypothesized Mean Difference is 0 and make sure you click Labels in the first row and click OK. You will get an output and this is the p-value you are looking for. P(T<=t) two-tail 0.2853 Question 12 1 / 1 point A researcher takes sample temperatures in Fahrenheit of 17 days from New York City and 18 days from Phoenix. Test the claim that the mean temperature in New York City is different from the mean temperature in Phoenix. Use a significance level of α=0.05. Assume the populations are approximately normally distributed with unequal variances. You obtain the following two samples of data. New York City Phoenix 99 94.2 95.5 72 93.2 86.8 102 122.1 85.4 114.4

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

80 94.7 86.4 89.7 75.4 104.7 79.5 77.6 83.4 106.8 64.3 98.6 65.5 91.5 87.7 82 104 97.7 74.3 64.9 59.5 82 82.8 72 116.2

The Hypotheses for this problem are: H 0 : μ 1 = μ 2 H 1 : μ 1 μ 2 Find the p-value. Round answer to 4 decimal places. Make sure you put the 0 in front of the decimal. p-value =___ ___ Answer: 0.0705 Hide question 12 feedback Use the Data Analysis Toolpak in Excel. Data - > Data Analysis -> scroll to where is says t:Test 2 Samples Assuming Unequal Variances -> OK Variable Input 1: is New York Variable Input 2: is Phoenix The Hypothesized Mean Difference is 0 and make sure you click Labels in the first row and click OK. You will get an output and this is the p-value you are looking for. P(T<=t) two-tail 0.0705 Question 13 1 / 1 point You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 30 night students, and the sample mean GPA is 2.35 with a standard deviation of 0.46. You sample 25 day students, and the sample mean GPA is 2.58 with a standard deviation of 0.47. Test the claim using a 5% level of significance. Assume the sample standard deviations are unequal and that GPAs are normally distributed. Hypotheses:

H 0 : μ 1 (?) μ 2 H 1 : μ 1 (?) μ 2 What are the correct hypotheses for this problem? H 0 : μ 1 ≠ μ 2 ; H 1 : μ 1 = μ 2 H 0 : μ 1 ≤ μ 2 ; H 1 : μ 1 ≥ μ 2 H 0 : μ 1 < μ 2 ; H 1 : μ 1 = μ 2 H 0 : μ 1 = μ 2 ; H 1 :μ 1 ≠ μ 2 H 0 : μ 1 = μ 2 ; H 1 : μ 1 > μ 2 H 0 : μ 1 ≥ μ 2 ; H 1 : μ 1 ≤ μ 2 Hide question 13 feedback This is a two tailed test because the keyword is different. This is a two tailed test because the keyword is different. Question 14 1 / 1 point In a two-tailed 2-sample z-test you find a P-Value of 0.0278. At what level of significance would you choose to reject the null hypothesis? Select all that apply. 0.05 0.08 0.10 0.01 Hide question 14 feedback If the p-value < alpha, then you Reject the Null and it is significant. 0.0278 < .05, .08 and .10.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

If the p-value < alpha, then you Reject the Null and it is significant. 0.0278 < .05, .08 and .10. If the p-value < alpha, then you Reject the Null and it is significant. 0.0278 < .05, .08 and .10. Question 15 1 / 1 point Which of the following is a requirement that must first be satisfied before running a z- test for the difference between two means? The standard deviations for both populations must be known. The standard deviations should come from each sample. The samples must be dependent. The populations must be approximately normal. Question 16 1 / 1 point In a random sample of 50 Americans five years ago (Group 1), the average credit card debt was $5,798. In a random sample of 50 Americans in the present day (Group 2), the average credit card debt is $6,511. Let the population standard deviation be $1,154 five years ago, and let the current population standard deviation be $1,645. Using a 0.01 level of significance, test if there is a difference in credit card debt today versus five years ago. What are the correct hypotheses for this problem? H 0 : μ 1 = μ 2 ; H 1 : μ 1 ≠ μ 2

H 0 : μ 1 ≤ μ 2 ; H 1 : μ 1 ≥ μ 2 H 0 : μ 1 ≠ μ 2 ; H 1 : μ 1 = μ 2 H 0 : μ 1 ≥ μ 2 ; H 1 : μ 1 ≤ μ 2 H 0 : μ 1 = μ 2 ; H 1 : μ 1 > μ 2 H 0 : μ 1 < μ 2 ; H 1 : μ 1 = μ 2 Hide question 16 feedback This is a two tailed test because of the keyword difference. Question 17 1 / 1 point A survey found that the average daily cost to rent a car in Los Angeles is $102.24 and in Las Vegas is $97.35. The data were collected from two random samples of 40 in each of the two cities and the population standard deviations are $5.98 for Los Angeles and $4.21 for Las Vegas. At the 0.05 level of significance, construct a confidence interval for the difference in the means and then decide if there is a significant difference in the rates between the two cities. Let the sample from Los Angeles be Group 1 and the sample from Las Vegas be Group 2. Confidence Interval (round to 4 decimal places): ___< μ 1 - μ 2 <___ ___ Answer for blank # 1: 2.6236 (50 %) Answer for blank # 2: 7.1564 (50 %) Hide question 17 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 LL = (102.24 - 97.35) - 1.96 *5.98240+4.21240 UL = (102.24 - 97.35) + 1.96 *5.98240+4.21240 Question 18 1 / 1 point

A 2011 survey, by the Bureau of Labor Statistics, reported that 91% of Americans have paid leave. In January 2012, a random survey of 1000 workers showed that 89% had paid leave. The resulting p-value is .0271; thus, the null hypothesis is rejected. It is concluded that there has been a decrease in the proportion of people, who have paid leave from 2011 to January 2012. What type of error is possible in this situation? Type I Error Standard Error Margin of Error Type II Error No error was made Question 19 1 / 1 point Match the symbol with the correct phrase. µ significance level confidence level parameter power P(Type II Error) Hide question 19 feedback This is a parameter of the population average.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Question 20 1 / 1 point A left-tailed z-test found a test statistic of z = -1.99. At a 5% level of significance, what would the correct decision be? Do not reject H 0 Do not reject H 1 Accept H 0 Reject H 1 Reject H 0 Hide question 20 feedback =NORM.S.DIST(-1.99,TRUE) = 0.023295468 < .05, Reject Ho.

Homework 6 Attempt 3

Related Documents