You may need to use the appropriate appendix table or technology to answer this question.

A simple random sample of 70 items resulted in a sample mean of 90. The population standard deviation is 10.

(a) Compute the 95% confidence interval for the population mean. Round your answers to two decimal places.

(b) Assume that the same sample mean was obtained from a sample of 140 items. Provide a 95% confidence interval for the population mean. Round your answers to two decimal places.

(c) What is the effect of a larger sample size on the interval estimate?

• A larger sample size provides a larger margin of error.
• A larger sample size does not change the margin of error.
• A larger sample size provides a smaller margin of error.

---

Answer:

Step 1 of 2

Given data:
The sample mean, $\overline{x}=90$
The population standard deviation, $\sigma =10$
The sample size, $n=70$
The confidence level = 95%

Explanation:
The sample data and population standard deviation are identified and depicted above along with the confidence level. The formula for the confidence interval is:

$CI=\overline{x}\pm z_c\times \frac{\sigma }{\sqrt{n}}$

---

Step 2 of 2

(a) For a 95% confidence level:

The significance level is calculated as:

$\alpha =1–0.95=0.05$

Critical value:

$z_{\alpha /2}=z_{0.025}$
Using standard Z-tables or Excel, $z_c\approx 1.96$

Now, we substitute these values into the confidence interval formula:

$CI=90\pm 1.96\times \frac{10}{\sqrt{70}}$

First, calculate the margin of error:

$1.96\times \frac{10}{\sqrt{70}}\approx 2.34$

Now, calculate the confidence interval:

$CI=90\pm 2.34$
$CI\approx (87.66,92.34)$

---

(b) For a larger sample size $n=140$:

Using the same steps as before:

$CI=90\pm 1.96\times \frac{10}{\sqrt{140}}$

First, calculate the margin of error:

$1.96\times \frac{10}{\sqrt{140}}\approx 1.66$

Now, calculate the confidence interval:

$CI=90\pm 1.66$
$CI\approx (88.34,91.66)$

---

(c) Explanation of margin of error:

The margin of error is inversely proportional to the square root of the sample size. This means that for larger sample sizes, the margin of error will be smaller.

The formula for the margin of error is:

$ME=z_c\times \frac{\sigma }{\sqrt{n}}$

---

Final Solution:

• (a) The 95% confidence interval with $n=70$ is $87.66$ to $92.34$.
• (b) The 95% confidence interval with $n=140$ is $88.34$ to $91.66$.
• (c) The larger the sample size, the smaller the margin of error.

---