Question

this problem requires PhStat solution.

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 7,500 hours. The population standard deviation is 1,000 hours. A random sample of 64 CFLs indicate a sample mean life of 7,250 hours.

1. At 0.05 level of significance, state your decision.

2. Using the critical value approach, is there evidence that the mean life is different from 7,500 hours?

3. Determine the p-value and interpret its meaning.

4. Construct and interpret a 95% confidence interval estimate of the population mean life of the CFLs.

Answer 1

Answer 2

1. At 0.05 level of significance, state your decision.

We will conduct a two-tailed hypothesis test at a 0.05 level of significance.

Null Hypothesis: The mean life of the CFLs is equal to 7,500 hours. Alternative Hypothesis: The mean life of the CFLs is not equal to 7,500 hours.

Let's denote the sample mean as x̄, the population standard deviation as σ, and the sample size as n. We have:

x̄ = 7,250 hours σ = 1,000 hours n = 64

We can use a z-test to test our hypothesis. The test statistic is given by:

z = (x̄ - μ) / (σ / sqrt(n))

where μ is the population mean.

Substituting the given values, we get:

z = (7,250 - 7,500) / (1,000 / sqrt(64)) z = -3.2

Using a standard normal distribution table, we find that the critical values for a two-tailed test at 0.05 level of significance are ±1.96. Since our test statistic falls outside this range, we reject the null hypothesis.

Therefore, we have sufficient evidence to conclude that the mean life of the CFLs is different from 7,500 hours.

2. Using the critical value approach, is there evidence that the mean life is different from 7,500 hours?

Yes, there is evidence that the mean life of the CFLs is different from 7,500 hours, since our calculated z-value (-3.2) falls outside the critical values of ±1.96.

3. Determine the p-value and interpret its meaning.

The p-value is the probability of obtaining a test statistic as extreme as, or more extreme than, the observed result under the null hypothesis. In this case, we have a two-tailed test, so we need to find the probability of obtaining a z-value less than -3.2 or greater than 3.2.

Using a standard normal distribution table, we find that the p-value is less than 0.01. This means that there is less than a 1% chance of obtaining a sample mean as far from the population mean as we observed, if the null hypothesis were true. Therefore, we reject the null hypothesis.

4. Construct and interpret a 95% confidence interval estimate of the population mean life of the CFLs.

We can construct a 95% confidence interval for the population mean life of the CFLs using the formula:

CI = x̄ ± z*(σ / sqrt(n))

where CI is the confidence interval, z is the critical value (1.96 for a 95% confidence interval), and the other variables have their previous values.

Substituting the given values, we get:

CI = 7,250 ± 1.96*(1,000 / sqrt(64)) CI = (6,944, 7,556)

We can be 95% confident that the true population mean life of the CFLs falls within this interval. This means that if we were to repeat this process many times, 95% of the resulting confidence intervals would contain the true population mean.