The quality-control manager at a compact fluorescent light bulb (CFL) factory ne

Question

The quality-control manager at a compact fluorescent light bulb (CFL) factory needs to determine whether the mean life of a large shipment of CFLs is equal to 7460 hours. The population standard deviation is 840 hours. A random sample of 64 light bulbs indicates a sample mean life of 7, 219 hours. a. At the 0.05 level of significance, is there evidence that the mean life is different from 7460 hours? b. Compute the p-value and interpret its meaning. c. Construct a 95% confidence interval estimate of the population mean life of the light bulbs. d. Compare the results of (a) and (c). What conclusions do you reach? e. Compare the results of parts (a) through (d) to those when the standard deviation is 1, 200 hours. a. Let mu be the population mean. Determine the null hypothesis. H_0, and the alternative hypothesis, H_1. What is/are the critical value(s)? What is the final conclusion? A. Reject H_0. There is not sufficient evidence to indicate that the mean life is different from 7460 hours. B. Fail to reject H_0. There is not sufficient evidence to indicate that the mean life is different from 7460 hours. C. Reject H _0 There is sufficient evidence to indicate that the mean life is different from 7460 hours. D. Fail to reject H_ o There is sufficient evidence to indicate that the mean life is different from 7460 hours. b. What is the p-value?

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: = 7460

Alternative hypothesis: 7460

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE) and the z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 105

z = (x - ) / SE

z = - 2.30

zcritical = + 1.96

Reject H0, There is sufficient evidence to indicate that mean is different from 7460.

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.

Since we have a two-tailed test, the P-value is the probability that the z statistic is less than - 2.30 or greater than 2.30.

We use the t Distribution Calculator to find P(t < - 2.30) = 0.0107, and P(t > 2.30) = 0.0107.

Thus, the P-value = 0.0107 + 0.0107 = 0.0214

Thus, the P-value = 0.021

Interpret results. Since the P-value (0.0214) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that mean life is different from 7460.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.