Question

Part IV: Using Minitab to perform hypothesis testing (or confidence intervals) for proportions based on two independent samples Note: Since our textbook is considering the analysis when the hypothesized difference is zero, BE SURE TO Use pooled estimate for the proportion for the Test Method!! Use Minitab output to answer the following problems [BE SURE TO ANSWER THE QUESTIONS] A Pew Internet and American Life Project report stated that 74% of 450 teenage boys surveyed posted their photo on their online profile, while 83% of the 500 teenage girls surveyed did so. a) At the 1% level of significance, do the data provide sufficient evidence to conclude that teenage boys post their photo online at a rate lower than teenage girls do? rates of online photo posting between teenage boys and girls. Interpret the confidence interval result.

Please solve Q#a and b using Minitab. List the step by step process, state the null hypothesis and the alternative hypothesis, and also state the conclusion in a clear sentence.

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P₁> P₂
Alternative hypothesis: P₁ < P₂

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)

p = 0.78737

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.02659
z = (p₁ - p₂) / SE

z = - 3.38

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than - 3.38.

Thus, the P-value = 0.0004.

Interpret results. Since the P-value (0.0004) is less than the significance level (0.01), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that teenage boys post their photo online at a rate lower than teenage girls.

b) 99% confidence interval for the difference between rates of online photo posting between teenage boys and girls is C.I = ( - 0.1585, - 0.0215).

$C.I = \left (p_{1}-p_{2} \right )\pm z_{\alpha /2}\times \sqrt{\frac{p_{1}(1-p_{1})} {n_{1}}+\frac{p_{2}(1-p_{2})}{n_{2}}}$

C.I = (0.74 - 0.83) + 2.576*0.02659

C.I = - 0.09 + 0.0685

C.I = ( - 0.1585, - 0.0215)