Question

A telemarketing company based in Utah has established a guideline stating that the average time for each completed call should be no more than 250 seconds. Recently the operations manager was concerned that calls were taking too long. The operations manager did not wish to assert that the calls were taking too long if the sample data did not strongly indicate this. A sample of 12 calls was selected and the following times (in seconds) were recorded. 194 278 302 140 245 234 268 208 302 190 320 255 Based on the sample data, develop appropriate null and alternate hypothesis and perform the relevant hypothesis test. Test the hypothesis at the 0.10 significance level using the p-value approach. (Assume that call times are normally distributed.). Report your results using 7-step method.

Answer 1

Before testing the hypothesis we need to calculate the mean and standard deviation of the given sample 194 278 302 140 245 234 268 208 302 190 320 255 as:

Mean = (194 + 278 + 302 + 140 + 245 + 234 + 268 + 208 + 302 + 190 + 320 + 255)/12
= 2936/12
Mean = 244.6667

and sample standard deviation as:

s = √(1/12 - 1) x ((194 - 244.6667)2 + ( 278 - 244.6667)2 + ( 302 - 244.6667)2 + ( 140 - 244.6667)2 + ( 245 - 244.6667)2 + ( 234 - 244.6667)2 + ( 268 - 244.6667)2 + ( 208 - 244.6667)2 + ( 302 - 244.6667)2 + ( 190 - 244.6667)2 + ( 320 - 244.6667)2 + ( 255 - 244.6667)2)
= √(1/11) x ((-50.6667)2 + (33.3333)2 + (57.3333)2 + (-104.6667)2 + (0.33330000000001)2 + (-10.6667)2 + (23.3333)2 + (-36.6667)2 + (57.3333)2 + (-54.6667)2 + (75.3333)2 + (10.3333)2)
= √(0.0909) x ((2567.11448889) + (1111.10888889) + (3287.10728889) + (10955.11808889) + (0.11108889000001) + (113.77848889) + (544.44288889) + (1344.44688889) + (3287.10728889) + (2988.44808889) + (5675.10608889) + (106.77708889))
= √(0.0909) x (31980.66666668)
= √(2907.0426000012)
= 53.9197

Now since the distribution is normal and based on the claim the hyppotheses are:

Step-1

Step-2

Type of Hypothesis:

Based on the hypothesis it will be a right-tailed test, but since the population standard deviation is not given hence t-distribution is applicable so, degree of freedom is used as df =n-1= 12-1=11

Step-3

Rejection region:

Based on the hypothesis and given significance level the critical score to test the hypothesis is computed using the excel formula for T-distribution which is =T.INV(1-0.10, 11), thus tc is computed as 1.3634.

so, reject Ho if t>tc

Step-4

Test Statistic:

Step-5

P-value:

The P-value for the calculated t-statistic is computed using excel formula for t-distribution which is =T.DIST.RT(-0.3426, 12), thus P-value computed as: 0.6311

Step-6

Decision:

Since t < tc hence we failed to reject the null hypothesis.

Step-7

Conclusion:

Since we failed to reject the null hypothesis and hence we conclude that there is insufficient evidence to support the claim.