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\\).
(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{1.73}\\).
(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 \\(\boxed{smaller}\\), since \\(\boxed{an increase}\\) in the sample size will \\(\boxed{decrease}\\) the standard error.
(c) if the confidence levels were 99.8% rather than 90%, would the margin of error be larger or smaller than the result in part (a)? explain.
the margin of error would be \\(\boxed{select}\\), since \\(\boxed{select}\\) in the confidence level will \\(\boxed{select}\\) the critical value.

Explanation:

Part (a)

Step 1: Recall the formula for confidence interval

The formula for a confidence interval for the population mean when the population standard deviation \(\sigma\) is known is \(\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\), 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, the significance level \(\alpha = 1 - 0.90 = 0.10\), so \(\alpha/2 = 0.05\). The critical value \(z_{0.05}\) (from the standard normal distribution table) is approximately 1.645.

Step 2: Identify the given values

We are given: \(\bar{x} = 1.63\), \(\sigma = 0.98\), \(n = 170\).

Step 3: Calculate the margin of error (E)

The margin of error \(E = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\). Plugging in the values:

\(E = 1.645 \times \frac{0.98}{\sqrt{170}}\)

First, calculate \(\sqrt{170} \approx 13.038\)

Then, \(\frac{0.98}{13.038} \approx 0.0752\)

Then, \(E = 1.645 \times 0.0752 \approx 0.1237\)

Step 4: Calculate the confidence interval

The lower limit is \(\bar{x} - E = 1.63 - 0.1237 \approx 1.5063 \approx 1.51\)

The upper limit is \(\bar{x} + E = 1.63 + 0.1237 \approx 1.7537 \approx 1.75\)

So the 90% confidence interval is \(1.51 < \mu < 1.75\) (which matches the given interval, so this checks out).

Part (b)

Step 1: Recall the margin of error formula

The margin of error \(E = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\). We can see that the margin of error is inversely proportional to the square root of the sample size (\(n\)). That is, as \(n\) increases, \(\frac{\sigma}{\sqrt{n}}\) decreases (since \(\sqrt{n}\) is in the denominator), and thus \(E\) decreases (because \(z_{\alpha/2}\) and \(\sigma\) are constant here).

Step 2: Analyze the change in sample size

The original sample size \(n = 170\), and the new sample size \(n = 195\) (which is larger than 170). Since \(n\) has increased, \(\sqrt{n}\) has increased, so \(\frac{\sigma}{\sqrt{n}}\) has decreased, and therefore the margin of error \(E\) will be smaller than the result in part (a). The reason is that an increase in the sample size will decrease the standard error (\(\frac{\sigma}{\sqrt{n}}\) is the standard error), and since the margin of error is based on the standard error, a smaller standard error leads to a smaller margin of error.

So the margin of error would be smaller, since an increase in the sample size will decrease the standard error.

Part (c)

Step 1: Recall the relationship between confidence level and critical value

The critical value \(z_{\alpha/2}\) (for a two - tailed test) increases as the confidence level increases. For a 90% confidence level, \(\alpha = 0.10\) and \(z_{\alpha/2}=1.645\). For a 99.8% confidence level, \(\alpha = 1 - 0.998=0.002\), so \(\alpha/2 = 0.001\). The critical value \(z_{0.001}\) (from the standard normal distribution table) is approximately 3.090, which is larger than 1.645.

Step 2: Recall the margin of error formula

The margin of error \(E = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\). Since the confidence level is increasing from 90% to 99.8%, the critical value \(z_{\alpha/2}\) is increasing (because a higher confidence level requires a larger critical value to capture more of the area under the normal curve). Since \(z_{\alpha/2}\) is in the numerator of the margin of error formula, and \(\sigma\) and \(n\) are constant (assuming the sample size and standard deviation don't change), an increase in \(z_{\alpha/2}\) will lead to an increase in the margin of error.

Also, the margin of error is related to the critical value: as the confidence level increases, the critical value increases, and since the margin of error is \(z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}\), the margin of error will be larger. The reason is that in the confidence level, the critical value will increase (a higher confidence level needs a larger critical value to be more certain of capturing the true mean), and since the margin of error depends on the critical value (along with standard deviation and sample size), a larger critical value leads to a larger margin of error.

So the margin of error would be larger, since an increase in the confidence level will increase the critical value.

Part (b) Answer:

smaller, increase, decrease

Part (c) Answer:

larger, increase, the margin of error is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\), as confidence level increases, \(z_{\alpha/2}\) (critical value) increases, so \(E\) increases. So the first select is "larger", the second select is "increase" (in confidence level leads to increase in critical value), the third select is "increase" (the margin of error will increase because critical value increases).

Answer:

Step 1: Recall the relationship between confidence level and critical value

The critical value \(z_{\alpha/2}\) (for a two - tailed test) increases as the confidence level increases. For a 90% confidence level, \(\alpha = 0.10\) and \(z_{\alpha/2}=1.645\). For a 99.8% confidence level, \(\alpha = 1 - 0.998=0.002\), so \(\alpha/2 = 0.001\). The critical value \(z_{0.001}\) (from the standard normal distribution table) is approximately 3.090, which is larger than 1.645.

Step 2: Recall the margin of error formula

The margin of error \(E = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\). Since the confidence level is increasing from 90% to 99.8%, the critical value \(z_{\alpha/2}\) is increasing (because a higher confidence level requires a larger critical value to capture more of the area under the normal curve). Since \(z_{\alpha/2}\) is in the numerator of the margin of error formula, and \(\sigma\) and \(n\) are constant (assuming the sample size and standard deviation don't change), an increase in \(z_{\alpha/2}\) will lead to an increase in the margin of error.

Also, the margin of error is related to the critical value: as the confidence level increases, the critical value increases, and since the margin of error is \(z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}\), the margin of error will be larger. The reason is that in the confidence level, the critical value will increase (a higher confidence level needs a larger critical value to be more certain of capturing the true mean), and since the margin of error depends on the critical value (along with standard deviation and sample size), a larger critical value leads to a larger margin of error.

So the margin of error would be larger, since an increase in the confidence level will increase the critical value.

Part (b) Answer:

smaller, increase, decrease

Part (c) Answer:

larger, increase, the margin of error is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\), as confidence level increases, \(z_{\alpha/2}\) (critical value) increases, so \(E\) increases. So the first select is "larger", the second select is "increase" (in confidence level leads to increase in critical value), the third select is "increase" (the margin of error will increase because critical value increases).