Question

g Delta Airlines is trying to determine if pilots are deliberately slowing down during a labor dispute. They know that their all their flights have a mean late time of 12.8 minutes with a standard deviation of 6.8 minutes. They took a random sample of 31 flights during the dispute and found they were 15.1 minutes late on average. Using a significance level of 0.05, is there any evidence to back the pilots claim that they are not slowing down?

Answer 1

`t=(15.1-12.8)/((6.8)/(√(31)))=1.883`

`p_v =P(t_(30)>1.883)=0.0347`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12.8 at 5% of signficance and the claim makes sense.

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=15.1` represent the sample mean

`\mu` population mean (variable of interest)

s=6.8 represent the sample standard deviation

n=31 represent the sample size

Solution to the problem

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the pilots claim that they are not slowing down, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 12.8`

Alternative hypothesis:
`\mu > 12.8`

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(15.1-12.8)/((6.8)/(√(31)))=1.883`

P-value

The degrees of freedom are given by:

`df = n-1=31-1=30`

Since is a right tailed test the p value would be:

`p_v =P(t_(30)>1.883)=0.0347`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12.8 at 5% of signficance and the claim makes sense.