Question

The new sheriff in town wants to reduce the average emergency response to under 30 mins which is the average response time of the current sheriff. Their are no past records so the standard deviation is not known before. The 1st week response times for the new sheriffs are

26,30,28,29,25,28,32,35,24,23

What kind of test is best fit for this problem?

B. One-sample t-test

C. Paired-samples t-test

D. Independent-samples t-test

What is the observed value and do you accept or reject null hypothesis?

Is it a one or two tailed test?

what is the effect size?

Did the new sheriff meet the goal, not meet the goal or surpass the goal?

Answer 1

What kind of test is the best fit for this problem?

B. One-sample t-test

Is it a one or two-tailed test?

One-tailed

The sample size is n = 10. The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:

	X	X2
	26	676
	30	900
	28	784
	29	841
	25	625
	28	784
	32	1024
	35	1225
	24	576
	23	529
Sum =	280	7964

The sample mean is computed as follows:

$\bar X = \frac{1}{n} \sum_{i=1}^n X_i = \frac{ 280}{ 10} = 28$

Also, the sample variance is

$s^2 = \frac{ 1}{n-1}\left(\sum_{i=1}^n X_i^2 - \frac{1}{n}\left(\sum_{i=1}^n X_i\right)^2 \right) = \frac{ 1}{ 10-1}\left( 7964 - \frac{ 280^2}{ 10} \right) = 13.778$

Therefore, the sample standard deviation s is

$s = \sqrt{s^2} = \sqrt{ 13.778} = 3.712$

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ ≤ 30

Ha: μ > 30

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, will be used.

(2) Rejection Region

Based on the information provided, the significance level is α = 0.05, and the critical value for a right-tailed test is t_c = 1.833.

(3) Test Statistics

The t-statistic is computed as follows:

$t = \frac{\bar X - \mu_0}{s/\sqrt{n}} = \frac{ 28 - 30}{ 3.712/\sqrt{ 10}} = -1.704$

(4) Decision about the null hypothesis

Since it is observed that t = -1.704 < t_c = 1.833, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.9387, and since p = 0.9387 > 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is greater than 30, at the 0.05 significance level.

Did the new sheriff meet the goal, not meet the goal or surpass the goal?

Hence, the mean time is less than 30 mins, or they meet the goal