Question

5. (12 pts) Using traditional methods it takes 98 hours to receive an advanced flying license. A new training technique using Computer Aided Instruction (CAI) has been proposed. A researcher used the technique on 140 students and observed that they an of 97 hours. Assume the population standard deviation is known to be 6. Is there evidence at the 0.1 level that the technique reduces the training time? Step 1. State the null and alternative hypotheses. Answer: Ho: Ha Step 2. Find the value of the test statistic. Round your answer to two decimal places. Answer: Step 3. Specify if the test is one-tailed or two-tailed. A) One-Tailed Test B) Two-Tailed Test Step 4. Find the P-value of the test statistic. Round your answer to four decimal places. Answer: Step 5. Identify the level of significance for the hypothesis test. Answer: Step 6. Make the decision to reject or fail to reject the null hypothesis. A) Reject Null Hypothesis B) Fail to Reject Null Hypothesis

Answer 1

- 1 Mo' l = 98 vs Hiiu98 2) Test statistics, Z=-1.972 3) It is one tail test 4) P-valule : 0.0243 5) level of significance, 2 - 0.10 6) P-value (0.0243) c & co.lo) Ho is Rejected. Reject Noll hypothers.

The provided sample mean is X = 97 and the known population standard deviation is o = 6, and the sample size is n=140. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Hou = 98 Hau <98 This corresponds to a left-tailed test for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.1, and the critical value for a left-tailed testis ze = -1.28. The rejection region for this left-tailed testis R={::%< -1.28} (3) Test Statistics The Z-statistic is computed as follows: X-MO o n 97-98 6/1/110 = -1.972 (4) Decision about the null hypothesis Since itis observed that := -1.972<%= -1.28, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0243, and since p = 0.0243 <0.1, it is concluded that the null hypothesis is rejected (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean uis less than 98, at the 0.1 significance level. Confidence Interval The 90% confidence interval is 96.166<u<97.834. Graphically Z-Test Results: t-stats = -1.972. p-value =0.0243 0.20 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

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