Question

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to determine of the mean number of unoccupied seats on all its flights is greater than 10. To accomplish this, the records of 60 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.2 seats and the sample standard deviation is 3.4 seats. Test the claim that mean number of unoccupied seats on all its flights is greater than 10 at the 5% significance level.

For any Hypothesis Test make sure to state Ho, Ha, Test statistic, p-value, whether you reject Ho, and your conclusion in the words of the claim.

For any confidence interval make sure that you interpret the interval in context, in addition to using it for inference.

round to the thousandths place

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 10
Alternative Hypothesis, Ha: μ > 10

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (11.2 - 10)/(3.4/sqrt(60))
t = 2.73

P-value Approach
P-value = 0.0042
As P-value < 0.05, reject the null hypothesis.

There is sufficient evidence to conclude that mean number of unoccupied seats on all its flights is greater than 10

sample mean, xbar = 11.2
sample standard deviation, s = 3.4
sample size, n = 60
degrees of freedom, df = n - 1 = 59

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2

ME = tc * s/sqrt(n)
ME = 2 * 3.4/sqrt(60)
ME = 0.878

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (11.2 - 2 * 3.4/sqrt(60) , 11.2 + 2 * 3.4/sqrt(60))
CI = (10.322 , 12.078)

we are 95% confident that the mean number of unoccupied seats on all its flights is between 10.322 and 12.078