In a recent poll, 793 adults were asked to identify their favorite seat when the

Question

In a recent poll, 793 adults were asked to identify their favorite seat when they fly, and 473 of them chose a window seat. Use a 0.01 significance level to test the claim that the majority of adults prefer window seats when they fly. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution. Which of the following is the hypothesis test to be conducted? H_0: p = 0.5 H_1: p 0.5 What is the test statistic? z = (Round to two decimal places as needed.) what is the P-value? P-value = (Round to three decimal places as needed.) What is the conclusion about the null hypothesis? Reject the null hypothesis because the P-value is lessthanorequalto the significance level, alpha. Reject the null hypothesis because the P-value is greater than the significance level, alpha. Fail to reject the null hypothesis because the P-value is lessthanorequalto the significance level, alpha. Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha What is the final conclusion?

Explanation / Answer

Solution:-

The correct hypothesis is (F), i.e.,

Null hypothesis: P = 0.50
Alternative hypothesis: P > 0.50

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too large.

For this analysis, the significance level is 0.01. The test method, shown in the next section, is a one-sample z-test.

Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

P = 473/793 = 0.60 (approximately)

= sqrt[ P * ( 1 - P ) / n ] = sqrt [(0.6 * 0.4) / 100] = sqrt(0.0024) = 0.04899
z = (p - P) / = (0.50 - 0.60) / 0.04899 = -2.04

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

We use the Normal Distribution Calculator to find P(z < -2.04)

The P-Value is 0.020675 or 0.021
The result is not significant at p < 0.01.

Conclusion. Fail to reject the null hypothesis because P-value is greater than the significance level alpha.

Interpret results. Since the P-value (0.021) is greater than the significance level (0.01), we can accept the null hypothesis.

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