Question

Boxes of cereal are labeled as containing 14 ounces. Following are the weights, in ounces, of a sample of 16 boxes. It is reasonable to assume that the population is approximately normal.
14.07
13.99
14.16
14.17
14.15
14.07
14.13
13.99
14.03
14.02
14.09
14.10
14.20
14.18
14.04
14.03
(a) Construct a 98% confidence interval for the mean weight. Round the answers to at least three decimal places.

Answer 1

To construct a 98% confidence interval for the mean weight of the cereal boxes, we can follow these steps:

1. Calculate the sample mean (\(\bar{x}\)).
2. Calculate the sample standard deviation (s).
3. Determine the critical value from the t-distribution for a 98% confidence level.
4. Use the formula for the confidence interval.

### Step 1: Calculate the Sample Mean

\[
\bar{x} = \frac{\sum x_i}{n}
\]

Where \(n\) is the number of boxes and \(x_i\) are the weights.

\[
\bar{x} = \frac{14.07 + 13.99 + 14.16 + 14.17 + 14.15 + 14.07 + 14.13 + 13.99 + 14.03 + 14.02 + 14.09 + 14.10 + 14.20 + 14.18 + 14.04 + 14.03}{16}
\]
\[
= \frac{226.33}{16} = 14.145625 \approx 14.146
\]

### Step 2: Calculate the Sample Standard Deviation

\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
\]

Calculating each \( (x_i - \bar{x})^2 \):

1. \( (14.07 - 14.146)^2 = 0.00576 \)
2. \( (13.99 - 14.146)^2 = 0.024576 \)
3. \( (14.16 - 14.146)^2 = 0.000196 \)
4. \( (14.17 - 14.146)^2 = 0.000576 \)
5. \( (14.15 - 14.146)^2 = 0.000016 \)
6. \( (14.07 - 14.146)^2 = 0.00576 \)
7. \( (14.13 - 14.146)^2 = 0.000256 \)
8. \( (13.99 - 14.146)^2 = 0.024576 \)
9. \( (14.03 - 14.146)^2 = 0.013056 \)
10. \( (14.02 - 14.146)^2 = 0.015876 \)
11. \( (14.09 - 14.146)^2 = 0.003136 \)
12. \( (14.10 - 14.146)^2 = 0.002116 \)
13. \( (14.20 - 14.146)^2 = 0.002976 \)
14. \( (14.18 - 14.146)^2 = 0.001176 \)
15. \( (14.04 - 14.146)^2 = 0.011556 \)
16. \( (14.03 - 14.146)^2 = 0.013056 \)

Now summing these squared differences:

\[
\sum (x_i - \bar{x})^2 = 0.024576 + 0.00576 + 0.000196 + 0.000576 + 0.000016 + 0.00576 + 0.000256 + 0.024576 + 0.013056 + 0.015876 + 0.003136 + 0.002116 + 0.002976 + 0.001176 + 0.011556 + 0.013056 \approx 0.16996
\]

Calculating the sample standard deviation:

\[
s = \sqrt{\frac{0.16996}{15}} \approx \sqrt{0.011331} \approx 0.1065
\]

### Step 3: Determine the Critical Value from the t-distribution

For a 98% confidence interval with \( n - 1 = 15 \) degrees of freedom, we check the t-distribution table or use a calculator.

The critical value \( t_{0.01, 15} \) (two-tailed) is approximately \( 2.602 \).

### Step 4: Calculate the Confidence Interval

The confidence interval is calculated as follows:

\[
\text{CI} = \bar{x} \pm t \cdot \frac{s}{\sqrt{n}}
\]

Calculating the standard error:

\[
\text{SE} = \frac{s}{\sqrt{n}} = \frac{0.1065}{\sqrt{16}} = \frac{0.1065}{4} = 0.026625
\]

Now, calculating the margin of error:

\[
ME = t \cdot \text{SE} = 2.602 \cdot 0.026625 \approx 0.0693
\]

Finally, the confidence interval:

\[
\text{CI} = 14.146 \pm 0.0693
\]

Thus:

\[
\text{Lower bound} = 14.146 - 0.0693 \approx 14.077
\]
\[
\text{Upper bound} = 14.146 + 0.0693 \approx 14.215
\]

### Final Result

The 98% confidence interval for the mean weight of the cereal boxes is approximately:

\[
(14.077, 14.215)
\]

Rounded to three decimal places, the interval is:

\[
(14.077, 14.215)
\]