Question

The sales representative for a manufacturer of a new product claims that the product will increase output per machine by more than 29 units per hour. A line manager installs the product on 20 of the machines, and finds that the average increase was 31 with a standard deviation of 6.2. Using ? ; 05, is there evidence that the population mean increase in output is greater than 29 units per hour? Do a complete and appropriate hypothesis test. Step 1 (Hypotheses) Ho: Click to select) Click to select) HA (Click to select) (Click to sel ect) Step 2.(Decision rule) Using only the appropriate statistical table in your textbook, the critical value for rejecting Ho is (Click to select) ' conventional rounding rules) eport your answer to 3 decimal places, using Step 3 (Test statistic) Using the sample data, the calculated value of the test statistic is (Click to select) report your answer to 2 decimal places, using conventional rounding rules) Step 4(Evaluate the null hypothesis) Should the null hypothesis be rejected? (Click to select) Step 5 (Practical conclusion) Should the line manager conclude that the population mean increase in output is greater than 29 units per hour? (Click to select Using only the appropriate statistical table in your textbook, what is the most accurate statement you can make about the numerical value of the p-value of this hypothesis test? Answer (provide a one-sentence statement about the p-value)

Answer 1

• Null hypothesis: ? <= 29

  Alternative hypothesis: ? > 29

  Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
• Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
• Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

  SE = s / sqrt(n) = 6.2 / sqrt(20) = 1.38636

  DF = n - 1 = 20 - 1 = 19

  t = (x - ?) / SE = (31 - 29)/1.38636 = 1.4426267

  where s is the standard deviation of the sample, x is the sample mean, ? is the hypothesized population mean, and n is the sample size.

  The observed sample mean produced a t statistic test statistic of 1.44. We use the t Distribution Calculator to find P-value.
• Thus, the P-Value is 0.082708.
• Interpret results. Since the P-value (0.082708) is greater than the significance level (0.05), we cannot reject the null hypothesis.