Question

19.

Suppose that in a random selection of 100 colored​ candies, 30​% of them are blue. The candy company claims that the percentage of blue candies is equal to 28​%. Use a 0.10 significance level to test that claim.

____________________

Identify the null and alternative hypotheses for this test. Choose the correct answer below.

A.

H0​:

p=0.28

H1​:

p<0.28

B.

H0​:

p≠0.28

H1​:

p=0.28

C.

H0​:

p=0.28

H1​:

p≠0.28

D.

H0​:

p=0.28

H1​:

p>

____________________

Identify the test statistic for this hypothesis test.

The test statistic for this hypothesis test is ​(Round to two decimal places as​ needed.)

Identify the​ P-value for this hypothesis test.

The​ P-value for this hypothesis test is ​(Round to three decimal places as​ needed.)

___________________

Identify the conclusion for this hypothesis test.

A.

Fail to reject

H0.

There

is

sufficient evidence to warrant rejection of the claim that the percentage of blue candies is equal to

28​%

B.

Reject

H0.

There

is not

sufficient evidence to warrant rejection of the claim that the percentage of blue candies is equal to

28​%

C.

Reject

H0.

There

is

sufficient evidence to warrant rejection of the claim that the percentage of blue candies is equal to

28​%

D.

Fail to reject

H0.

There

is not

sufficient evidence to warrant rejection of the claim that the percentage of blue candies is equal to

28​%

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.28

H_a : p 0.28

= 0.30

P₀ = 0.28

1 - P₀ = 1 - 0.28 = 0.72

Test statistic = z =

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.30 - 0.28 / [(0.28 * 0.72) / 100]

Test statistic = z = 0.45

P(z > 0.45 ) = 1 - P(z < 0.45) = 1 - 0.674 = 0.326

P-value = 2 * P(z > 0.45 )

P-value = 2 * 0.326

P-value = 0.652

= 0.10   

P-value >

Fail to reject the null hypothesis

D.Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the percentage of blue candies is equal to 28​%