Question

A study was done to determine whether the proportion of younger adults who subscribe to a traditional TV service (Comcast, DirecTV, Dish, etc.) differs according to age. The results are summarized in the following table:

	Age
Traditional TV?	>= 20 and < 30	>= 30 and < 40	Total
Yes	257	276	533
No	743	615	1358
Total	1000	891	1891

1. (a) At the .05 level of significance, use a 2-test to see if the sample data provides evidence that the proportion who subscribe to traditional TV is different for the two age groups considered. Perform your test using both the critical value and p-value approach.

2. (b) When only two proportions are involved, as is the case here, there is an equivalent two-sided z-test for testing the question posed in Part (a). This test is presented in text Section 10.3 (optional reading, shouldn’t be necessary to answer this question).Use software to generate the z-statistic for comparing two proportions and perform the test using both the critical value and p-value approach. Use the version of the test in Minitab (see ‘Options/Test Method’) that works with a ‘pooled estimate’ ofthe population proportion.

   Then verify that the 2-stat from Part (a) is the square of the z-stat (give or taking rounding), that the critical 2 value from Part (a) is the square of the critical z value, and that both statistics have the same p-values.

Answer 1

(a) Answer: Date: ---31/5/2019 Expected frequencies will be calculated as follows: ath row total jth row to 20 grand total Following table shows the expected frequencies: >= 20 and < 30 | >= 30 and < 40 | Total Yes 281.861 251.139 533 No 718.139 639.861 1358 891 Total1000 1891 Following table shows the calculations for chi square test statistics: O-E)*2/E 257 281.861 2.192816037 743718.139 0.860654164 276 251.139 2.461064673 615 639.861 0.965943105 Total 6.48047798 Following is the test statistics:

Following is the test statistics: (0-E)--6.48 Degree of freedom: df (number of rows -1) (number of columns-1 (2-1)(2-1)-1 The p-value is 0.0109 Since p-value is less than 0.05 so we reject the null hypothesis. The critical value is: 3.841 SinceX > 3.841 so we reject the null hypothesis. Excel function used for p-value:"-CHIDIST(6.48,1)" Excel function used for critical-value:"-CHIINV(0.05,1) Here we have x-257, n1-1000, pi = X1/n =0.257 ^276,n2-891p,/n -0.3098 Hypotheses are:

Level of significance: α-005 Test is two tailed. The pooled proportion is: (257+276)/(1000+891)-0.2819 两:/ǐy, The standard error is: 'SE-11p(1-2)|-+-| = 0.0207 n2 Test Statistics: z= (A-A)-(A-P.)--2.55 SE Critical value: CV = 1.96 Rejection Region: If z<- 1.96 or z > 1.96, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis. P-value 0.0108 Decision: Since p-value is less than level of significance so we reject the null hypothesis

Decision: Since p-value is less than level of significance so we reject the null hypothesis. Excel function for critical value: NORMSINV(1-(0.05/2)) Excel function for p-value: 2*(1-NORMSDIST(2.55)) Both tests give same results and same p value.