Question

The US Department of Agriculture estimates that 80% of American families cook dinner nightly. In a recent poll of 289 American families, 76.125% families said they cook dinner nightly. Use a 0.025 significance level to test the claim that the proportion of American families who cook dinner nightly is smaller than 80%.

The test statistic is: z=

The Critical Value is: zα2=

Answer 1

Solution :

This is the left tailed test .

The null and alternative hypothesis is

H₀ : p 0.80

H_a : p < 0.80

= 0.76125

P₀ = 0.80

1 - P₀ = 1 -- 0.80 = 0.20

Test statistic = z =

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.76125 - 0.80 / [(0.80 * 0.20) / 289]

Test statistic = z = -1.65

= 0.025    

= P(Z < -1.96 ) = 0.025  

z = -1.96

test statistic > critical value

Reject the null hypothesis .