Question

In​ 1997, a survey of 860 households showed that 161 of them use​ e-mail. Use those sample results to test the claim that more than​ 15% of households use​ e-mail. Use a 0.05 significance level.

What is the P-Value?

Answer 1

Solution :

Given that,

= 0.15

1 - = 0.85

n = 860

x = 161

Level of significance = = 0.05

Point estimate = sample proportion = = x / n = 0.187

This a right (One) tailed test.

The null and alternative hypothesis is,

Ho: p = 0.15

Ha: p 0.15

Test statistics

z = ( - ) / *(1-) / n

= ( 0.187 - 0.15) / (0.15*0.85) /860

= 3.056

P-value = P(Z>z)

= 1 - P(Z <z )

= 1- P(Z < 3.056)

= 1 - 0.9989

= 0.0011

The p-value is p = 0.0011, and since p = 0.0011 < 0.05, it is concluded that the null hypothesis is rejected.