Question

A publisher reports that 64% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 130 found that 60% of the readers owned a particular make of car. Is there sufficient evidence at the 0.05 level to support the executive's claim?

State the null and alternative hypotheses.

Find the value of the test statistic. Round your answer to two decimal places.

Specify if the test is one-tailed or two-tailed.

Determine the P-value of the test statistic. Round your answer to four decimal places.

Identify the value of the level of significance.

Make the decision to reject or fail to reject the null hypothesis.

State the conclusion of the hypothesis test.

Answer #1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H_{0} : p = 0.64

H_{a} : p
0.64

= 0.60

Test statistic = z

=
- P_{0} / [P_{0
*} (1 - P_{0} ) / n]

= 0.60 - 0.64/ [(0.60 * 0.40) / 130]

= -0.93

P(z < -0.93) = 0.1762

P-value = 0.3524

level of significance = = 0.05

P-value >

Fail to reject the null hypothesis .

There is not sufficient evidence to suggest that the percentage is actually different from the reported percentage .

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