Question

In a simple random sample of 27 companies, 25 have both Twitter and Facebook accounts. Can we conclude that more than 82% of companies have both Twitter and Facebook accounts?

What is the alternative hypothesis? ˆ p > 0.82 ˆ p ≥ 0.82 ˆ p > 0.93 p > 0.93 p ≥ 0.93 ˆ p ≥ 0.93 p ≥ 0.82 p > 0.82

Choose the correct probability statment for the p − value. P ( Z ≥ 0.93 − 0.82 √ 0.93 ⋅ 0.07 27)

P ( Z ≥ 0.93 − 0.82 √ 0.82 ⋅ 0.18 27 )

P ( Z ≥ 0.82 − 0.93 √ 0.93 ⋅ 0.07 27 )

P ( Z ≥ 0.82 − 0.93 √ 0.82 ⋅ 0.18 27 )

State your decision given a p − value of 0.12 using a reasonable significance level.

Fail to reject H 0 Accept H 0 Reject H 0

The correct conclusion is that we ______ conclude that the percent of _______ companies that have both Twitter and Facebook accounts ________%.

Answer 1

Solution:
Our claim is that more than 82% of companies have both Twitter and Facebook accounts, So null and alternative hypothesis can be written as
Null hypothesis H0: p = 0.82
Alternate hypothesis Ha: p > 0.82
Solution(b)
Sample size = 27
Sample proportion p^ = 25/27 = 0.93
Z test stat can be calculated as
Z- test statistic = (p^ - p)/sqrt(p*(1-p)/n) = (0.93-0.82)/sqrt(0.82*(1-0.82)/27)
so probability statement can be written as
P(Z>=(0.93-0.82)/sqrt(0.82*0.18/27)), So its correct answer is B.
Here Also given that p-value = 0.12
So at alpha = 0.05, we are failed to reject the null hypothesis as the p-value is greater than alpha.
So the correct conclusion is that we can not conclude that more than 82% of companies have both Twitter and Facebook accounts.