Questions: Here are summary statistics for the weights of Pepsi in randomly selected cans: n=36, x̄=0.82409 lb, s=0.00567 lb. Use a confidence level of 95% to complete parts (a) through (d) below. a. Identify the critical value tα/2 used for finding the margin of error. tα/2=2.03 (Round to two decimal places as needed.) b. Find the margin of error. E=0.00192 lb (Round to five decimal places as needed.) c. Find the confidence interval estimate of μ. 0.82217 lb<μ<0.82601 lb (Round to five decimal places as needed.)

Solution Steps

The critical value \( t_{\alpha/2} \) for a 95% confidence level is given as: \[ t_{\alpha/2} = 2.03 \]

The margin of error \( E \) is calculated using the formula: \[ E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \] Substituting the values: \[ E = 2.03 \cdot \frac{0.00567}{\sqrt{36}} = 0.00185 \] Thus, the margin of error is: \[ E = 0.00185 \text{ lb} \]

The confidence interval for the mean \( \mu \) is given by: \[ \bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \] Calculating the lower and upper bounds: \[ \text{Lower Bound} = 0.82409 - 2.03 \cdot \frac{0.00567}{\sqrt{36}} = 0.82224 \] \[ \text{Upper Bound} = 0.82409 + 2.03 \cdot \frac{0.00567}{\sqrt{36}} = 0.82594 \] Thus, the confidence interval is: \[ 0.82224 \text{ lb} < \mu < 0.82594 \text{ lb} \]

Final Answer