A publisher reports that 47%47% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 280280 found that 43%43% of the readers owned a personal computer. Make the decision to reject or fail to reject the null hypothesis at the 0.010.01 level.

The null hypothesis was not rejected.  

The proportion of readers who own a personal computer is 47%.

Step-by-step explanation:

The claim made by a publisher is that 47% of  their readers own a personal computer.

A single proportion z-test can be used to determine whether the claim made by the publisher is authentic or not.

The hypothesis for this test can be defined as follows:

H₀: The proportion of readers who own a personal computer is 47%, i.e. p = 0.47.

Hₐ: The proportion of readers who own a personal computer is different from 47%, i.e. p ≠ 0.47.

The information provided is:

The test statistic is:

The test statistic value is, z = -1.34.

Decision rule:

If the p-value of the test is less than the significance level α = 0.01 then the null hypothesis will be rejected and vice-versa.

Compute the p-value as follows:

*Use a z table for the probability.

The p-value of the test is 0.18.

p-value = 0.18 > α = 0.01

The null hypothesis was failed to be rejected at 1% level of significance.

Conclusion:

There is enough evidence to support the claim made by the publisher. Hence, it can be concluded that the proportion of readers who own a personal computer is 47%.

3/6   and 9/18

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