Q. The following are the times between six calls for an ambulance in a certain city and the patient’s arrival at the hospital: $27, 15, 20, 32, 18$ and $26$ minutes. Use these figures to judge the reasonableness of the ambulance service’s claim that it takes on the average 20 minutes between the call for an ambulance and the patient’s arrival at the hospital.
My attempt:
Here $\bar{x}=23$ and $s=6.39$. Assume $\mu=20$ is the normal population mean, the test statistic value is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}=1.15$. The entry in t-Table for $\alpha =0.1$ and $df=5$ is $1.476$. Now, how to proceed with the p value?
The $p$-value for a two-sided test is simply $$2 \Pr[T_5 > 1.15] \approx 0.30,$$ where $T_5$ is a Student's $t$ random variable with $5$ degrees of freedom. Since an ambulance arriving sooner than the claimed average time is not a detriment to the hospital's reputation, the one-sided $p$-value is simply $0.15$. This does not furnish strong evidence that the true average response time exceeds $20$ minutes.