SOLUTION: 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 for

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: 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 for      Log On

 Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Algebra: Probability and statistics Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Probability-and-statistics 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(57290)   (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 ----------------------------------------- Ho: u(daily) = 53 Ha: u(daily) > 53 --- 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. ================================================== Cheers, Stan H.