Question

A school was attempting to determine the proportion of students who have access to the internet in their homes. Demographic data suggests that 85% of students will have access to the internet at home, but this school believes the proportion at their school may be lower. A simple random sample of 120 students is selected, and of those 120, 95 students report having access to the internet at home. If a hypothesis test were to be performed, what would be the correct test statistic and P-value

Answer 1

Answer:

Z = -1.789

P value = 0.036807

Step-by-step explanation:

Here

H0: p = 0.85 students will have access to the internet at home

Ha: p < 0.85 students will have access to the internet at home

As per the given data, out of 120 sampled students, 95 have access to internet at home

This is 79.17 % of the total number of sampled students.

The Z values is given by

$Z = \frac{p-p'}{\sqrt{\frac{p(1-p)}{n} } }$

Substituting the given values, we get -

$Z = \frac{0.7917-0.85}{\sqrt{\frac{0.85(1-0.85)}{120} } }\\Z = -1.789$

The sample proportion of 0.7917 is about -1.789 standard errors below the population proportion given in the null hypothesis.

Considering, result is significant at p < .05, the P-Value is .036807.