Questions: Here are summary statistics for the weights of Pepsi in randomly selected cans: n=36, x̄=0.82414 lb, s=0.00573 lb. Use a confidence level of 95% to complete parts (a) through (d) below. a. Identify the critical value t(α / 2) used for finding the margin of error. t(α / 2)=2.03 (Round to two decimal places as needed.)

Solution Steps

We have the following summary statistics for the weights of Pepsi in randomly selected cans:

• Sample size \( n = 36 \)
• Sample mean \( \bar{x} = 0.82414 \, \text{lb} \)
• Sample standard deviation \( s = 0.00573 \, \text{lb} \)
• Confidence level \( 95\% \)

To calculate the confidence interval for the mean, we use the formula: \[ \bar{x} \pm z \frac{s}{\sqrt{n}} \] For a \( 95\% \) confidence level, the critical value \( z \) is approximately \( 1.96 \). Thus, we compute: \[ \text{Margin of Error} = z \frac{s}{\sqrt{n}} = 1.96 \cdot \frac{0.00573}{\sqrt{36}} = 1.96 \cdot \frac{0.00573}{6} \approx 0.00187 \] The confidence interval is then: \[ (0.82414 - 0.00187, 0.82414 + 0.00187) = (0.82227, 0.82601) \] Rounding to two decimal places, we have: \[ \text{Confidence Interval} = (0.82, 0.83) \]

The critical value \( t_{\alpha/2} \) is calculated based on the margin of error and the sample standard deviation: \[ t_{\alpha/2} = \frac{\text{Margin of Error}}{\frac{s}{\sqrt{n}}} \] Using the margin of error calculated earlier: \[ t_{\alpha/2} = \frac{0.00187}{\frac{0.00573}{\sqrt{36}}} = \frac{0.00187}{0.000955} \approx 5.24 \]

Final Answer