A food snack manufacturer samples 11 bags of pretzels off the assembly line and weighs their contents. If the sample mean is 15.2 oz. and the sample standard deviation is
0)50 oz., find the 95% confidence interval of the true mean.

Question

Quizplus · Accepted Answer

To find the 95% confidence interval for the true mean, we use the formula for the confidence interval of the mean with a t-distribution (since the population standard deviation is unknown and the sample size is small): $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $t_{\alpha/2}$ is the t-score corresponding to the desired confidence level and degrees of freedom (n-1). Given $\bar{x} = 15.2$ oz, $s = 0.50$ oz, $n = 11$, and for a 95% confidence level with 10 degrees of freedom, the t-score is approximately 2.228 (from t-distribution tables). Plugging these values into the formula gives a margin of error of $2.228 \cdot \frac{0.50}{\sqrt{11}} \approx 0.336$. Thus, the 95% confidence interval is $15.2 \pm 0.336$, which is approximately $14.9 < \mu < 15.5$.

A Food Snack Manufacturer Samples 11 Bags of Pretzels Off 15.0<μ<15.415.0 < \mu < 15.415.0<μ<15.4

A Food Snack Manufacturer Samples 11 Bags of Pretzels Off $15.0 < \mu < 15.4$