Question

Are U.S. adults aged 18-29 more likely to use Facebook than U.S. adults aged 30-49?” The Pew Research Center conducted a survey with a random sample of U.S. adults in January 2018.2 Out of 351 adults aged 18-29, 282 used Facebook. Out of 525 adults aged 30-49, 410 used Facebook.

a. Compute the value of the statistic of interest.

b. Are the appropriate conditions met to conduct a hypothesis test using the z-distribution? Explain.

c. Conduct a hypothesis test to answer the first research question. Use a significance level of 0.05.

Compute test statistic:

P-value from the appropriate theoretical distribution StatKey =

Formal decision and why:

Conclusion:

Answer 1

a) x1 = 282

n1 = 351

x2 = 410

n1 = 525

$\hat{p2}=\frac{x2}{n2}=\frac{410}{525}=0.7810$

Test statistic is

$z=\frac{\hat{p1}-\hat{p2}}{\sqrt{p*(1-p)\left ( \frac{1}{n1} +\frac{1}{n2}\right)}}$

$p=\frac{x1+x2}{n1+n2}$

$p=\frac{282+410}{351+525} =\frac{692}{876}=0.7900$

$z=\frac{0.8034-0.7810}{\sqrt{0.79*(1-0.79)\left ( \frac{1}{351} +\frac{1}{525}\right)}} =\frac{0.0225}{\sqrt{0.1659*0.004754}}=0.80$

b) Conditions:

$n1*\hat{p1}$ = 351*0.8034 = 282 > 10

$n1*(1-\hat{p1})$ = 351*0.1966 = 69 > 10

$n2*\hat{p2}$ = 525*0.7810 = 410 > 10

$n2*(1-\hat{p2})$ = 525*0.2190 = 115 > 10

So all conditions are met we can use z distribution.

c) Claim: U.S. adults aged 18-29 more likely to use Facebook than U.S. adults aged 30-49.

The null and alternative hypothesis is

H0: P1 = P2

H1: P1 > P2

Level of significance = 0.05

P-value = P(Z > 0.80) = 0.2119

P-value > 0.05 we fail to reject null hypothesis.

Conclusion: U.S. adults aged 18-29 NOT more likely to use Facebook than U.S. adults aged 30-49