Question 187151: I NEED HELP IN STATISTICS. PLEASE HELP. THANK YOU SO MUCH.
Quality Polls contends that an agent conducts a mean of 53 in-depth home surveys every week. A streamlined survey form has been introduced, and the pollster wants to evaluate its effectiveness. A random sample of 17 agents showed their total number of surveys conducted during a week.
58
58
63
48
51
54
59
57
56
59
53
51
53
54
64
59
60
At the .05 level of significance, can we conclude that the mean number of interviews conducted by the agents is more than 53 per week? Why? Explain your decision on the null hypothesis.
Hypothesis Test: Mean vs. Hypothesized Value
53.000 hypothesized value
56.882 mean Data
3.199 std. dev.
0.776 std. error
17 n
16 df
5.00 t
.0001 p-value (one-tailed, upper)
55.237 confidence interval 95.% lower
58.527 confidence interval 95.% upper
1.645 half-width
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! At the .05 level of significance, can we conclude that the mean number of interviews conducted by the agents is more than 53 per week? Why? Explain your decision on the null hypothesis.
Hypothesis Test: Mean vs. Hypothesized Value
53.000 hypothesized value
56.882 mean Data
3.199 std. dev.
0.776 std. error
17 n
16 df
5.00 t
.0001 p-value (one-tailed, upper)
55.237 confidence interval 95.% lower
58.527 confidence interval 95.% upper
1.645 half-width
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Ho: u(daily) = 53
Ha: u(daily) > 53
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Conclusion: since the p-value is less than 5%, reject Ho.
The test provided evidence that the agent's daily average is greater than 53.
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Cheers,
Stan H.
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