Question

A publisher reports that 62% 62 % of their readers own a laptop. A marketing executive wants to test the claim that the percentage is actually more than the reported percentage. A random sample of 130 130 found that 70% 70 % of the readers owned a laptop. Is there sufficient evidence at the 0.10 0.10 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

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.62
Alternative Hypothesis, Ha: p > 0.62

This is right tailed test

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.7 - 0.62)/sqrt(0.62*(1-0.62)/130)
z = 1.88

P-value Approach
P-value = 0.0301

0.1 is the levlof significance

As P-value < 0.1, reject the null hypothesis.

There is sufficient evidence to conclude that the percentage is actually more than the reported percentage.