Question

Suppose that in a random selection of 100 colored candies, 28 % of them are blue. The candy company claims that the percentage of blue candies is equal to 22%. Use a 0.05 significance level to test that claim. identify the original claim identify the null and the alternative hypothesis identify the test statistic. identify the P-value Do we reject or fail-to-reject the null hypothesis. Write down your final conclusion that addresses the original claim

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.22

H_a : p 0.22

n = 100

= 0.28

P₀ = 0.22

1 - P₀ = 1 - 0.22 = 0.78

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.28 - 0.22 / [(0.22 * 0.78) / 100]

= 1.45

Test statistic = 1.45

P(z > 1.45) = 1 - P(z < 1.45) = 1 - 0.9265 = 0.0735

P-value = 2 * P(z > 1.45) = 2 * 0.0735 = 0.147

P-value = 0.1470

= 0.05

P-value >

Fail to reject the null hypothesis .

There is no sufficinet evidence to support the claims that the percentage of blue candies is equal to 22% .