To construct a 95% confidence interval for the true population mean textbook weight, follow these steps:

1. Identify the sample mean ($\bar{x}$), population standard deviation ($\sigma$), and sample size ($n$):
- $\bar{x} = 72$ ounces
- $\sigma = 8.9$ ounces
- $n = 42$

2. Determine the critical value for a 95% confidence interval using the z-distribution. For a 95% confidence level, the critical value ($z^*$) is approximately 1.96.

3. Calculate the standard error of the mean (SE):
$$
SE = \frac{\sigma}{\sqrt{n}} = \frac{8.9}{\sqrt{42}} \approx 1.374
$$

4. Compute the margin of error (ME):
$$
ME = z^* \times SE = 1.96 \times 1.374 \approx 2.69
$$

5. Determine the confidence interval:
$$
\text{Lower limit} = \bar{x} - ME = 72 - 2.69 = 69.31
$$
$$
\text{Upper limit} = \bar{x} + ME = 72 + 2.69 = 74.69
$$

Therefore, the 95% confidence interval for the true population mean textbook weight is $[69.31, 74.69]$ ounces.

Question

To construct a 95% confidence interval for the true population mean textbook weight, follow these steps:

1. Identify the sample mean ($\bar{x}$), population standard deviation ($\sigma$), and sample size ($n$):
   - $\bar{x} = 72$ ounces
   - $\sigma = 8.9$ ounces
   - $n = 42$

2. Determine the critical value for a 95% confidence interval using the z-distribution. For a 95% confidence level, the critical value ($z^*$) is approximately 1.96.

3. Calculate the standard error of the mean (SE):
   $$
   SE = \frac{\sigma}{\sqrt{n}} = \frac{8.9}{\sqrt{42}} \approx 1.374
   $$

4. Compute the margin of error (ME):
   $$
   ME = z^* \times SE = 1.96 \times 1.374 \approx 2.69
   $$

5. Determine the confidence interval:
   $$
   \text{Lower limit} = \bar{x} - ME = 72 - 2.69 = 69.31
   $$
   $$
   \text{Upper limit} = \bar{x} + ME = 72 + 2.69 = 74.69
   $$

Therefore, the 95% confidence interval for the true population mean textbook weight is $[69.31, 74.69]$ ounces.

Knowee AI · Accepted Answer

To construct a 95% confidence interval for the true population mean textbook weight, follow these steps:

1. Identify the sample mean ($\bar{x}$), population standard deviation ($\sigma$), and sample size ($n$):
   - $\bar{x} = 72$ ounces
   - $\sigma = 8.9$ ounces
   - $n = 42$

2. Determine the critical value for a 95% confidence interval using the z-distribution. For a 95% confidence level, the critical value ($z^*$) is approximately 1.96.

3. Calculate the standard error of the mean (SE):
   $$
   SE = \frac{\sigma}{\sqrt{n}} = \frac{8.9}{\sqrt{42}} \approx 1.374
   $$

4. Compute the margin of error (ME):
   $$
   ME = z^* \times SE = 1.96 \times 1.374 \approx 2.69
   $$

5. Determine the confidence interval:
   $$
   \text{Lower limit} = \bar{x} - ME = 72 - 2.69 = 69.31
   $$
   $$
   \text{Upper limit} = \bar{x} + ME = 72 + 2.69 = 74.69
   $$

Therefore, the 95% confidence interval for the true population mean textbook weight is $[69.31, 74.69]$ ounces.

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