Question

8. Hypothesis Testing: One or Two Means, Various Samples (a) A sample of 40 light bulbs from Company #1 have a mean lifetime of 650 hours with a standard deviation of 25 hours. They are advertized to have a 750 hour life. Is the mean lifetime of this sample inconsistent with the advertized life? Test the alt hypo (Ha: u < 750) against the null hypo (Ho: u = 750) at the 0.05 significance level. (6) A sample of 40 light bulbs from Company #1 have a mean lifetime of 650 hours with a standard deviation of 25 hours; a sample of 30 from Company #2 have a mean of 640 hours with a std dev of 10 hours. (Assume unequal std deviations: s1 s2) Are Bulbs #1 really better? Test the alt hypo (Ha: ul > 2) against the null hypo (Ho: ul = u2) at the 0.01 significance level. (c) A sample of 20 light bulbs from Company #1 have a mean lifetime of 650 hours with a standard deviation of 25 hours; a sample of 20 from Company #2 have a mean of 640 hours with a std dev of 20 hours. (Assume equal std deviations: s1 = s2) Are Bulbs #1 really better? Test the alt hypo (Ha: ul >u2) against the null hypo (Ho: ul = u2) at the 0.01 significance level. (d) Five people who use lights differently tested pairs of bulbs from Companies #1 and #2. The table below gives the lifetimes they observed. Are Bulbs #1 really better? Test the alt hypo (Ha: ul > 2) against the null hypo (Ho: ul - 2) at the 0.01 significance level. Abe # 1 642 #2 630 | Diff | 12 Betty 645 634 11 Curt 650 640 10 Dave Ellen 655 658 645 651 | 10T7 Mean 650 640 1 0

Answer 1

Excel add-in (PH Stat) procedure: Step 1: In EXCEL, Select Add-Ins > PH Stat > One-Sample Tests > t Test for the mean, Sigma unknown. Step 2: Enter 750 under Null hypothesis. Step 3: Enter 0.05 under Level of significance. Step 4: In Sample Statistic Options, choose Sample Statistic Known and enter 40 as the Sample size and 650 as the Sample mean, and 25 as the Sample standard deviation. Step 5: In Test Options, choose Lower Tail Test. Step 6: Click OK Excel add-in (PH Stat) output:

t Test for Hypothesis of the Mean 0 Data Null Hypothesis = Level of significance 1 Sample Size Sample Mean Sample Standard Deviation L 750 .05 40 650 25 Intermediate Calculations Standard Error of the Mean 3.9528 Degrees of Freedom t Test Statistic -25.2982 Lower-Tail Test Lower Critical Value p-Value Reject the null hypothesis -1.6849 0.0000 The value of the test statistic is t = -25.2982 The critical value is -1.6849. P-value = 0.

Conclusion: Use a significance level, a = 0.05. Here, p-value is 0, which is less than the value of a = 0.05. That is, p-value =0) sal=0.05). Therefore, by the rejection rule, reject the null hypothesis H. Interpretation: There is sufficient evidence at level of significance a = 0.05. That is, we can conclude that the population mean less than 750.