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 σ=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 1.51 < μ < 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 smaller, since an increase in the sample size will decrease the standard error.
part: 2 / 4
part 3 of 4
(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 larger, since an increase in the confidence level will increase the critical value.
part: 3 / 4
part 4 of 4
(d) based on the confidence interval constructed in part (a), is it likely that the mean number of personal computers is less than 1.88?
it select likely that the mean number of personal computers is less than 1.88.

Explanation:

Part 1 of 4 (a)

Step1: Recall Confidence Interval Formula

For a population mean with known standard deviation, the confidence interval is $\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$. Here, $\bar{x} = 1.63$, $\sigma = 0.98$, $n = 170$, and for 90% confidence, $z_{\alpha/2} \approx 1.645$.

Step2: Calculate Margin of Error (ME)

$ME = z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = 1.645 \times \frac{0.98}{\sqrt{170}} \approx 1.645 \times 0.0753 \approx 0.1239$.

Step3: Find Confidence Interval

Lower bound: $\bar{x} - ME = 1.63 - 0.1239 \approx 1.51$. Upper bound: $\bar{x} + ME = 1.63 + 0.1239 \approx 1.75$.

Margin of error formula is $ME = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$. As sample size $n$ increases, $\frac{\sigma}{\sqrt{n}}$ (standard error) decreases. So with $n = 195 > 170$, ME is smaller.

Margin of error depends on $z_{\alpha/2}$. Higher confidence level (99.8% > 90%) means larger $z_{\alpha/2}$, so ME increases.

Answer:

$1.51 < \mu < 1.75$

Part 2 of 4 (b)