Question

BlueSky Air claims that at least 80% of its flights arrive on time. A random sample of 160 BlueSky Air flights revealed that 115 arrive on time. Do the data provide sufficient evidence to contradict the claim by BlueSky Air (i.e., you would like to see whether the percentage of the airline's flight is below what the airline claims)?

Part iii) Compute the P-value (please round to four decimal places):

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P > 0.80
Alternative hypothesis: P < 0.80

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

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

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

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.03162
z = (p - P) / S.D

z = - 2.57

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

Since we have a one-tailed test, the P-value is the probability that the z-score is less than -2.57.

Thus, the P-value = 0.0051

Interpret results. Since the P-value (0.0051) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that that at least 80% of its flights arrive on time.