Question

You want to know if there's a difference between the proportions of high-school students and college students who read newspapers regularly. Out of a random sample of 500 high-school students, 287 say they read newspapers regularly, and out of a random sample of 420 college students, 252 say they read newspapers regularly. For this question, think of high-school students as sample one and college students as sample two.
A. Construct a 95% confidence interval for the difference between the proportions of high-school students and college students who read newspapers regularly. Be sure to show that you've satisfied the conditions for using a z-interval. (5 points)

B. Draw a conclusion, based on your 95% confidence interval, about the difference between the two proportions. (2 points)

C. If you wanted to use a test statistic to determine whether the proportion of high-school students who read newspapers regularly is significantly lower than the proportion of college students who read newspapers regularly, what would you use as your null and alternative hypotheses? (2 points)

D. Calculate p ˆ, the pooled estimate of the population proportions you'd use for a significance test about the difference between the proportions of high-school students and college students who read newspapers regularly. (1 point)

E. Demonstrate that these samples meet the requirements for using a zprocedure for a significance test about the difference between two proportions. (2 points)

F. Calculate SEp ˆ , the pooled estimate of the standard errors of the proportions you'd use in a z-procedure for a significance test about the difference between two proportions. (1 point)

G. Calculate your test statistic and P-value for the hypothesis test H0 : p1 = p2 , Ha : p1 < p2 . (4 points)

H. Draw a conclusion about the difference between the two proportions using α = .05. Is the proportion of high-school students who read the newspaper on a regular basis less than the proportion of college students who read newspapers regularly?

Answer 1

Answer:

The proportion of high-school students who read the newspaper on a regular basis is not  less than the proportion of college students who read newspapers regularly

Step-by-step explanation:

Sample                       High school           college         total

    N                               500                      420              920

    X                                287                      252              539

    p                                 0.574                     0.6            0.586

A) Sample size is very large and also proportions are nearer to 0.5 hence binomial approximates to normal so Z can be used.

Var (p1-p2) = Var(p1)+Var(p2)

Std def for difference = $\sqrt{p(1-p)(\frac{1}{n_1} +\frac{1}{n_2})} \\=\sqrt{0.586(1-0.586)(\frac{1}{500} +\frac{1}{420})\\=0.0326$

Margin of error =1.96* std error = 0.0639

Confidence interval = p difference ±margin of error

= (-0.0899, -0.0639)

B) Since 95% confidence interval contains 0, there is no significant difference between the two proportions

C) H0: p1 = p2

Ha: p1 <p2

D) Pooled estimate = 0.586

E) Yes because sample sizes are large and proportion is nearer to 0.5

G) Z = -0.7975

p value = 0.21186(one tailed)

H) Since p value is greater than 0.05 our significant level, we accept null hypothesis.

The proportion of high-school students who read the newspaper on a regular basis is not  less than the proportion of college students who read newspapers regularly