A publisher reports that $39 \%$ 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 400 found that $45 \%$ of the readers owned a laptop. Is there sufficient evidence at the 0.05 level to support the executive's claim?

Step 4 of 7 : Determine the $P$-value of the test statistic. Round your answer to four decimal places.

Step 1 ：Given that the publisher reports that 39% 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 400 found that 45% of the readers owned a laptop. We are asked to determine if there is sufficient evidence at the 0.05 level to support the executive's claim.

Step 2 ：The null hypothesis is that the true proportion of readers who own a laptop is 39%, and the alternative hypothesis is that the true proportion is more than 39%.

Step 3 ：We can calculate the P-value using a one-sample z-test for proportions. The test statistic is given by the formula: \(z = \frac{{p_{hat} - p_0}}{{\sqrt{{\frac{{p_0 * (1 - p_0)}}{n}}}}}\) where \(p_{hat}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and n is the sample size.

Step 4 ：Substituting the given values into the formula, we get \(z = \frac{{0.45 - 0.39}}{{\sqrt{{\frac{{0.39 * (1 - 0.39)}}{400}}}}} = 2.460277104314189\).

Step 5 ：The P-value is then the area to the right of the test statistic on the standard normal distribution, because we are testing for a greater proportion. The P-value is approximately 0.0069.

Step 6 ：Since the P-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that there is sufficient evidence to support the executive's claim that the true proportion of readers who own a laptop is more than 39%.

Step 7 ：Final Answer: The P-value is approximately \(\boxed{0.0069}\).