Question

According to a study in a previous year, 61.0 % of households nationwide used natural gas for heating during a year. Recently, a survey of 2,800 randomly selected households showed that 62.0 % used natural gas. Use a 0.05 significance level to test the claim that the 61.0 % national rate has changed.

what is the z score?

what is the p value?

state the conclusion

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H0 : p = 0.61

Ha : p 0.61

= 0.62

P0 = 0.61

1 - P0 = 1 - 0.61 = 0.39

Test statistic = z =

= - P0 / [P0 * (1 - P0 ) / n]

= 0.62 - 0.61/ [(0.61 * 0.39) / 2800]

Test statistic = z = 1.08

P(z > 1.08) = 1 - P(z < 1.08) = 1 - 0.8599 = 0.1401

P-value = 2 * P(z > 1.08)

P-value = 2 * 0.1401

P-value = 0.2802

= 0.05    

P-value >

Fail to reject the null hypothesis.