Question

a sample of college seniors showed that 15 of 50 who owned playstation 3 spent more than an hour a day playing games, compared with 35 of 100 who owned zoo box.

a) test the claim that p1 < p2 using a 0.01 significance level.

Test statistic:

p value:

conclusion about the claim

b) constuct a 95% confidence interval for the difference between the two population proportions

Answer 1

For Play station we have that the sample size is N1​=50, the number of favorable cases is X1​=15, so then the sample proportion is p^​1​=X1/N1​​​=15/50​=0.3

For zoo box, we have that the sample size is N2=100, the number of favorable cases is X2=35, so then the sample proportion is p^2 = X2/N2= 35/100= 0.35
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The value of the pooled proportion is computed as

Also, the given significance level is α=0.01.

The following null and alternative hypotheses need to be tested:

Ho:p1​=p2​

Ha:p1​>p2​

This corresponds to a right-tailed test, for which a z-test for two population proportions needs to be conducted.

The z-statistic is computed as follows

Using the P-value approach: The p-value is p=0.7299, and since p=0.7299≥0.01, it is concluded that the null hypothesis is not rejected.

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 who owned playstation 3 spent more than an hour a day playing games, compared with who owned zoo box., at the 0.01 significance level.

b) The 95% confidence interval for difference of proportion is −0.208<p1−p2 <0.108.