Question

Hugger Polls contends that an agent conducts a mean of 53 in-depth home surveys every week. A streamlined survey form has been introduced, and Hugger wants to evaluate its effectiveness. The number of in-depth surveys conducted during a week by a random sample of 15 agents are: 53 57 50 55 58 54 60 52 59 62 60 60 51 59 56 Click here for the Excel Data File At the 0.05 level of significance, can we conclude that the mean number of interviews conducted by the agents is more than 53 per week? (Round your answer to 3 decimal places.) Compute the value of the test statistic. (Round your answer to 3 decimal places.) What is your decision regarding H0? Reject Do not reject Estimate the p-value.

Answer 1

:-  Average number of interviews conducted by the agents

Number	Values (X)
1	53	11.56
2	57	0.36
3	50	40.96
4	55	1.96
5	58	2.56
6	54	5.76
7	60	12.96
8	52	19.36
9	59	6.76
10	62	31.36
11	60	12.96
12	60	12.96
13	51	29.16
14	59	6.76
15	56	0.16
Total	846	195.6

Mean

Standard deviaiton

To Test :-

H₀ :-  

H₁ :-  

Test Statistic :-

t = 3.523

Test Criteria :-

Reject null hypothesis if

= 3.523 > 1.761, We reject null hypothesis

Conclusion :- Accept Altervative hypothesis

Decision based on P value

t = 3.523

looking for the t = 3.523 across n-1 degree of freedom i . e 15 - 1 = 14 degree of freedom for one tail

3.523 lies between 2.977 and 3.787 repspective P value is 0.005 and 0.001

Using excel we can calculate exact P value = 0.00168

Reject null hypothesis if P value < = 0.05 level of significance

0.00168 < 0.05, we reject null hypothesis

Conclusion :- Accept Alternative Hypothesis

There is sufficient evidence to conclude that there is statistically significant in the average number of interviews conducted by the agents more than 53 per week.