Question

1. a random sample of 56 fluorescent light bulbs has a mean life of 645 hours with a standard deviation of 31 hours. construct a 95% confidence interval for the population mean.
2. a candidate for governor of a particular state claims to be favored by at least half of the voters. write the

Explanation:

Step1: Identify the formula for confidence interval

For a sample mean, when the population standard deviation \(\sigma\) is unknown (we use sample standard deviation \(s\) instead) and the sample size \(n\) is large (\(n \geq 30\)), we use the z - interval (by the Central Limit Theorem). The formula for the confidence interval for the population mean \(\mu\) is:
\(\bar{x}\pm z_{\alpha/2}\frac{s}{\sqrt{n}}\)
where \(\bar{x}\) is the sample mean, \(z_{\alpha/2}\) is the z - score corresponding to the level of confidence, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step2: Determine the values of the parameters

We are given that:

• The sample size \(n = 56\)
• The sample mean \(\bar{x}=645\) hours
• The sample standard deviation \(s = 31\) hours
• For a 95% confidence interval, the significance level \(\alpha=1 - 0.95 = 0.05\). Then \(\alpha/2=0.025\). The \(z\) - score \(z_{0.025}\) (the z - score such that the area to the right of it is 0.025) is approximately 1.96 (from the standard normal distribution table).

Step3: Calculate the margin of error

The margin of error \(E=z_{\alpha/2}\frac{s}{\sqrt{n}}\)
Substitute the values: \(n = 56\), \(s = 31\), \(z_{\alpha/2}=1.96\)
First, calculate \(\sqrt{n}=\sqrt{56}\approx7.4833\)
Then, \(\frac{s}{\sqrt{n}}=\frac{31}{7.4833}\approx4.1425\)
Then, \(E = 1.96\times4.1425\approx8.12\)

Step4: Calculate the confidence interval

The lower limit of the confidence interval is \(\bar{x}-E=645 - 8.12 = 636.88\)
The upper limit of the confidence interval is \(\bar{x}+E=645 + 8.12 = 653.12\)

Answer:

The 95% confidence interval for the population mean is \((636.88, 653.12)\) (or in the form \(645\pm8.12\))