Question

A marketing firm claims the proportion of people who use social networks every day is different for people with no college than for people who are college graduates.

To test the claim they conduct surveys of both groups asking “Do you use a social network like Facebook or Instagram every day?”, the results appear below.

No College: X1 = 55 n1 = 83

College Graduates: X2 = 74    n2 = 97

a) Identify the Null and Alternative Hypotheses for the marketing firm’s claim.

b) For a 0.05 significance level find the critical value for the test.

c) Calculate the test statistic.

Round off all proportions to 3 significant digits.

Round off your test statistic to two decimal places.

d) Calculate the p-value for the test.

e) What is your decision about the null hypothesis?

Answer 1

Two-Proportion Z test

The following information is provided: (a) Sample 1 - The sample size is N1 = 83, the number of favorable cases is X1 = 55 and the sample proportion is p^1​=X1/N1​=55/83​=0.6627 (b) Sample 2 - The sample size is N2 = 97, the number of favorable cases is X2 = 74 and the sample proportion is p^2​=X2/N2​=74/97​=0.7629 and the significance level is α=0.05 Pooled Proportion The value of the pooled proportion is computed as (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p1 = p2 Ha: p1 ≠ p2 This corresponds to a Two-tailed test, for which a z-test for two population proportions needs to be conducted. (2a) Critical Value Based on the information provided, the significance level is α=0.05, therefore the critical value for this Two-tailed test is Zc​=1.96. This can be found by either using excel or the Z distribution table. (2b) Rejection Region The rejection region for this Two-tailed test is |Z|>1.96 i.e. Z>1.96 or Z<-1.96 (3) Test Statistics The z-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is p =P(|Z|>1.4877)=0.1368 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that |Z|=1.4877 < Zc​=1.96, it is then concluded that the null hypothesis is Not rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.1368, and since p=0.1368>0.05, it is concluded that the null hypothesis is Not rejected. (6) Conclusion It is concluded that the null hypothesis Ho is Not rejected. Therefore, there is Not enough evidence to claim that the population proportion p1 is different than p2, at the 0.05 significance level.

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