a.
Find the value of the
a.

Answer to Problem 20E
The value of chi-square statistic is 0.965.
Explanation of Solution
Calculation:
The
Contingency table:
A contingency table is obtained as using two qualitative variables. One of the qualitative variable is row variable that has one category for each row of the table another is column variable has one category for each column of the table.
The hypotheses are:
Null Hypothesis:
Alternate Hypothesis:
Now, it is obtained that,
Gender | Male | Female | Row Total |
Optimistic | 1,148 | 815 | 1,963 |
Pessimistic | 225 | 178 | 403 |
Column Total | 1,373 | 993 | 2,366 |
Expected frequencies:
The expected frequencies in case of contingency table is obtained as,
Now, using the formula of expected frequency it is found that the expected frequency for the optimistic male is obtained as,
Hence, in similar way the expected frequencies are obtained as,
Gender | Male | Female |
Optimistic | ||
Pessimistic |
Chi-Square statistic:
The chi-square statistic is obtained as
The accept and reject can be rewritten as,
Male | 1 |
Female | 2 |
Test Statistic:
Software procedure:
Step -by-step software procedure to obtain test statistic using MINITAB software is as follows:
- Select Stat > Table > Cross Tabulation and Chi-Square.
- Check the box of Raw data (categorical variables).
- Under For rows enter Row.
- Under For columns enter Gender.
- Check the box of Count under Display.
- Under Chi-Square, click the box of Chi-Square test.
- Select OK.
- Output using MINITAB software is given below:
Thus, the value of chi-square statistic is 0.965.
b.
Find the proportion of men who were optimistic.
b.

Answer to Problem 20E
The proportion of men who were optimistic is 0.836.
Explanation of Solution
Calculation:
From part (a), it is found that,
Gender | Male | Female | Row Total |
Optimistic | 1,148 | 815 | 1,963 |
Pessimistic | 225 | 178 | 403 |
Column Total | 1,373 | 993 | 2,366 |
Hence, the proportion of men who were optimistic is,
Thus, the proportion of men who were optimistic is 0.836.
c.
Find the proportion of women who were optimistic.
c.

Answer to Problem 20E
The proportion of women who were optimistic is 0.821.
Explanation of Solution
Calculation:
From part (a), it is found that,
Gender | Male | Female | Row Total |
Optimistic | 1,148 | 815 | 1,963 |
Pessimistic | 225 | 178 | 403 |
Column Total | 1,373 | 993 | 2,366 |
Hence, the proportion of women who were optimistic is,
Thus, the proportion of women who were optimistic is 0.821.
d.
Find the test statistic z for testing the null hypothesis that the two proportions are equal versus the alternative that they are not equal.
d.

Answer to Problem 20E
The test statistic z for testing the null hypothesis that the two proportions are equal versus the alternative that they are not equal is 0.9821.
Explanation of Solution
Calculation:
Assume that
It is also assumed that
The random variables
The assumptions for performing a Hypothesis Test for the difference between two population proportions are defined as,
- The two random samples are independent to each other.
- Each population size is at least 20 times of the sample size.
- The individuals in the each sample are divided into two categories.
- The minimum sample size in each category is 10.
A random sample of 1,373 men and another random sample of 993 women are asked in the General Survey that whether they were optimistic about the future. There is no statistical relationship between these two samples. Hence, the two random samples are independent to each other.
The number of men and women in United States are much larger than the drawn samples. Hence, population size is more than 20 times of the sample size
The individuals in the each sample are classified in two categories. One is optimistic and another is pessimistic.
The sample size of men is 1,148 and the sample size of women is 815.
Now, as all the assumptions for performing a Hypothesis Test for the difference between two populations proportions are satisfied, then one can proceed to perform a Hypothesis Test for the difference between two population proportions.
The hypotheses are:
Null Hypothesis:
That is, the population proportions of men and proportion of women who were optimistic are same.
Alternative Hypothesis:
That is, the population proportions of men and proportion of women who were optimistic are not same.
The test statistic z is defend as
The pooled proportion is defined as
From part (a), it is found that,
Gender | Male | Female | Row Total |
Optimistic | 1,148 | 815 | 1,963 |
Pessimistic | 225 | 178 | 403 |
Column Total | 1,373 | 993 | 2,366 |
Thus,
From part (b), it is found that,
Hence, the proportion of men who were optimistic is 0.836.
Therefore,
From part (c) it is found that proportion of women who were optimistic is 0.821.
Therefore,
Hence,
Thus, the test statistic value is,
e.
Prove that
e.

Explanation of Solution
Calculation:
From part (a), it is found that
Hence,
From part (a), it is found that value of chi-square statistic is 0.965.
Hence, it is proved that
f.
Find the P-values for each of these tests using technology.
Prove that P-values are equal.
f.

Answer to Problem 20E
The P-values for each of these tests is 0.326.
Explanation of Solution
Calculation:
From part (a), it is found that the P-value for the chi square test is 0.326.
Using z test:
Software procedure:
Step by step procedure to obtain the P-values using the MINITAB software is given below:
- Choose Stat > Basic Statistics > 2 Proportions.
- Choose Summarized data.
- In First sample, enter Trials as 1,373 and
Events as 1,148. - In Second sample, enter Trials as 993 and Events as 815.
- Check Perform hypothesis test.
- Under Test Method choose Use the pooled estimate of the proportion.
- Click OK.
The output is MINITAB software is given below:
Therefore, the P-value is 0.326.
Hence, it is proved that the P-values for each of these tests are same.
g.
Conclude that when a contingency table has two rows and two columns, the chi square tests equivalent to the test for the difference between proportions.
g.

Explanation of Solution
It is found that the P-values for chi-square test and z test are same. Moreover, the square of z test statistic is same as the chi-square test statistic.
Hence, it can be concluded that when a contingency table has two rows and two columns, the chi square tests equivalent to the test for the difference between proportions.
Want to see more full solutions like this?
Chapter 10 Solutions
ALEKS 360 ESSENT. STAT ACCESS CARD
- Consider the hypothesis test Ho: = 622 against H₁: 6 > 62. Suppose that the sample sizes are n₁ = 20 and n₂ = 8, and that = 4.5; s=2.3. Use a = 0.01. (a) Test the hypothesis. Round your answers to two decimal places (e.g. 98.76). The test statistic is fo = i The critical value is f = Conclusion: i the null hypothesis at a = 0.01. (b) Construct the confidence interval on 02/022 which can be used to test the hypothesis: (Round your answer to two decimal places (e.g. 98.76).) iarrow_forward2011 listing by carmax of the ages and prices of various corollas in a ceratin regionarrow_forwardس 11/ أ . اذا كانت 1 + x) = 2 x 3 + 2 x 2 + x) هي متعددة حدود محسوبة باستخدام طريقة الفروقات المنتهية (finite differences) من جدول البيانات التالي للدالة (f(x . احسب قيمة . ( 2 درجة ) xi k=0 k=1 k=2 k=3 0 3 1 2 2 2 3 αarrow_forward
- 1. Differentiate between discrete and continuous random variables, providing examples for each type. 2. Consider a discrete random variable representing the number of patients visiting a clinic each day. The probabilities for the number of visits are as follows: 0 visits: P(0) = 0.2 1 visit: P(1) = 0.3 2 visits: P(2) = 0.5 Using this information, calculate the expected value (mean) of the number of patient visits per day. Show all your workings clearly. Rubric to follow Definition of Random variables ( clearly and accurately differentiate between discrete and continuous random variables with appropriate examples for each) Identification of discrete random variable (correctly identifies "number of patient visits" as a discrete random variable and explains reasoning clearly.) Calculation of probabilities (uses the probabilities correctly in the calculation, showing all steps clearly and logically) Expected value calculation (calculate the expected value (mean)…arrow_forwardif the b coloumn of a z table disappeared what would be used to determine b column probabilitiesarrow_forwardConstruct a model of population flow between metropolitan and nonmetropolitan areas of a given country, given that their respective populations in 2015 were 263 million and 45 million. The probabilities are given by the following matrix. (from) (to) metro nonmetro 0.99 0.02 metro 0.01 0.98 nonmetro Predict the population distributions of metropolitan and nonmetropolitan areas for the years 2016 through 2020 (in millions, to four decimal places). (Let x, through x5 represent the years 2016 through 2020, respectively.) x₁ = x2 X3 261.27 46.73 11 259.59 48.41 11 257.96 50.04 11 256.39 51.61 11 tarrow_forward
- If the average price of a new one family home is $246,300 with a standard deviation of $15,000 find the minimum and maximum prices of the houses that a contractor will build to satisfy 88% of the market valuearrow_forward21. ANALYSIS OF LAST DIGITS Heights of statistics students were obtained by the author as part of an experiment conducted for class. The last digits of those heights are listed below. Construct a frequency distribution with 10 classes. Based on the distribution, do the heights appear to be reported or actually measured? Does there appear to be a gap in the frequencies and, if so, how might that gap be explained? What do you know about the accuracy of the results? 3 4 555 0 0 0 0 0 0 0 0 0 1 1 23 3 5 5 5 5 5 5 5 5 5 5 5 5 6 6 8 8 8 9arrow_forwardA side view of a recycling bin lid is diagramed below where two panels come together at a right angle. 45 in 24 in Width? — Given this information, how wide is the recycling bin in inches?arrow_forward
- 1 No. 2 3 4 Binomial Prob. X n P Answer 5 6 4 7 8 9 10 12345678 8 3 4 2 2552 10 0.7 0.233 0.3 0.132 7 0.6 0.290 20 0.02 0.053 150 1000 0.15 0.035 8 7 10 0.7 0.383 11 9 3 5 0.3 0.132 12 10 4 7 0.6 0.290 13 Poisson Probability 14 X lambda Answer 18 4 19 20 21 22 23 9 15 16 17 3 1234567829 3 2 0.180 2 1.5 0.251 12 10 0.095 5 3 0.101 7 4 0.060 3 2 0.180 2 1.5 0.251 24 10 12 10 0.095arrow_forwardstep by step on Microssoft on how to put this in excel and the answers please Find binomial probability if: x = 8, n = 10, p = 0.7 x= 3, n=5, p = 0.3 x = 4, n=7, p = 0.6 Quality Control: A factory produces light bulbs with a 2% defect rate. If a random sample of 20 bulbs is tested, what is the probability that exactly 2 bulbs are defective? (hint: p=2% or 0.02; x =2, n=20; use the same logic for the following problems) Marketing Campaign: A marketing company sends out 1,000 promotional emails. The probability of any email being opened is 0.15. What is the probability that exactly 150 emails will be opened? (hint: total emails or n=1000, x =150) Customer Satisfaction: A survey shows that 70% of customers are satisfied with a new product. Out of 10 randomly selected customers, what is the probability that at least 8 are satisfied? (hint: One of the keyword in this question is “at least 8”, it is not “exactly 8”, the correct formula for this should be = 1- (binom.dist(7, 10, 0.7,…arrow_forwardKate, Luke, Mary and Nancy are sharing a cake. The cake had previously been divided into four slices (s1, s2, s3 and s4). What is an example of fair division of the cake S1 S2 S3 S4 Kate $4.00 $6.00 $6.00 $4.00 Luke $5.30 $5.00 $5.25 $5.45 Mary $4.25 $4.50 $3.50 $3.75 Nancy $6.00 $4.00 $4.00 $6.00arrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman





