Question

Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes. A chi-square test was performed. Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade the p-value.

Answer 1

The p-value will be "0.0549".

Explanation:

The given values are:

Time, t = 15 minutes

Df, σ = 25-1 = 24

Now,

⇒
`H_(0):\sigma^2\leq 150`

and,

⇒
`H_(1):\sigma^2>150`

As we know,

Chi square =
`((n-1)s^2)/((\sigma^2))`

On putting the values in the above formula, we get

⇒ =
`(24* 15^2)/(150)`

⇒ =
`(24* 225)/(150)`

⇒ =
`36`

Therefore, p-value = 0.0549

The p-value determined > 0.05, the null hypothesis also isn't dismissed at point 0.05.