Question

how many computers? in a simple random sample of 170 households, the sample mean number of personal computers was 1.63. assume the population standard deviation is \\(\sigma = 0.98\\).
part: 0 / 4
part 1 of 4
(a) construct a 90% confidence interval for the mean number of personal computers. round the answer to at least two decimal places.
a 90% confidence interval for the mean number of personal computers is \\(\boxed{1.51} < \mu < \boxed{1.75}\\).
part: 1 / 4
part 2 of 4
(b) if the sample size were 195 rather than 170, would the margin of error be larger or smaller than the result in part (a)? explain.
the margin of error would be \\(\text{select}\\), since \\(\text{select}\\) in the sample size will \\(\text{select}\\) the standard error.

Explanation:

Step1: Recall Confidence Interval Formula

The formula for a confidence interval for the population mean (when population standard deviation \(\sigma\) is known) is \(\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z_{\alpha/2}\) is the critical value, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. For a 90% confidence interval, \(\alpha = 1 - 0.90 = 0.10\), so \(\alpha/2 = 0.05\). The \(z\)-score for \(z_{0.05}\) is approximately \(1.645\) (from standard normal distribution tables).

Step2: Identify Given Values

We have \(\bar{x} = 1.63\), \(\sigma = 0.98\), \(n = 170\), and \(z_{\alpha/2} = 1.645\).

Step3: Calculate the Margin of Error (E)

First, calculate the standard error \(\frac{\sigma}{\sqrt{n}}=\frac{0.98}{\sqrt{170}}\). Let's compute \(\sqrt{170}\approx13.038\), so \(\frac{0.98}{13.038}\approx0.0752\). Then, the margin of error \(E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} = 1.645 \cdot 0.0752\approx0.1237\).

Step4: Construct the Confidence Interval

The lower bound is \(\bar{x} - E = 1.63 - 0.1237\approx1.5063\approx1.51\) (rounded to two decimal places). The upper bound is \(\bar{x} + E = 1.63 + 0.1237\approx1.7537\approx1.75\) (rounded to two decimal places).

For part (b): The margin of error formula is \(E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\). Since \(n\) is in the denominator of \(\frac{\sigma}{\sqrt{n}}\), as \(n\) increases (from 170 to 195), \(\frac{\sigma}{\sqrt{n}}\) decreases, so the margin of error \(E\) will be smaller. This is because a larger sample size provides more information, reducing the uncertainty (margin of error). The margin of error is related to the standard error (\(\frac{\sigma}{\sqrt{n}}\)) by the critical value, so as the sample size increases, the standard error decreases, leading to a smaller margin of error.

Answer:

(a) The 90% confidence interval is \(1.51 < \mu < 1.75\).
(b) The margin of error would be smaller, since an increase in the sample size will decrease the standard error.