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Accordingto statistics released by the Department of Civil Aviation, the average delayof flights is equal to 16.8 minutes, flight delays are assumed to follow anormal distribution.  However, from asample of 50 flights, the average delay was estimated to be 19.5 minutes andthe sample standard deviation was 6.6 minutes.Conducta hypothesis test to disprove the claim that the average delay is equal to 16.8minutes at a= 0.01.

Question

Accordingto statistics released by the Department of Civil Aviation, the average delayof flights is equal to 16.8 minutes, flight delays are assumed to follow anormal distribution.  However, from asample of 50 flights, the average delay was estimated to be 19.5 minutes andthe sample standard deviation was 6.6 minutes.Conducta hypothesis test to disprove the claim that the average delay is equal to 16.8minutes at a= 0.01.

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Solution

To conduct a hypothesis test, we first need to set up our null and alternative hypotheses.

The null hypothesis (H0) is that the average delay is equal to 16.8 minutes. H0: μ = 16.8

The alternative hypothesis (H1) is that the average delay is not equal to 16.8 minutes. H1: μ ≠ 16.8

We are given that the sample mean (x̄) is 19.5 minutes, the sample standard deviation (s) is 6.6 minutes, and the sample size (n) is 50.

We can use these values to calculate the test statistic (z) using the formula:

z = (x̄ - μ) / (s/√n)

Substituting the given values:

z = (19.5 - 16.8) / (6.6/√50) = 2.58

Now, we need to find the critical z-value for a two-tailed test at α = 0.01. Looking this up in a z-table, we find that the critical z-values are -2.58 and +2.58.

Since our calculated z-value of 2.58 is greater than the critical z-value of +2.58, we reject the null hypothesis.

Therefore, we have enough evidence to disprove the claim that the average delay is equal to 16.8 minutes at a significance level of 0.01.

This problem has been solved

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