Question

You are the product manager for Joe Totter Supermarket. You are interested in validating the contents of the one-pound bags of Keen potato chips. You select a random sample of 9 bags and empty the contents of each bag, one at a time, onto a scale and record the weights in pounds. Below are the recorded sample weights:
1.11.21.00.91.11.31.21.1

Assuming the weights of potato chips are normally distributed, estimate the mean weight of a bag of Keen potato chips using a 95% confidence interval.

Answer 1

Answer:

Hence, the mean of the potato chips lie between the interval $(1.006, 1.194)$

Given :

data is :

  $1.1,1.2,1.0,0.9,1.1,1.3,1.2,1.1,1.0$

Confidence level is $95\%$.

To find :

Confidence interval.

Explanation :

Mean $\bar{x}$ $=\frac{\sum x}{n}$

  $\Rightarrow \bar{x}=\frac{1.1+1.2+1.0+0.9+1.1+1.3+1.2+1.1+1.0}{9}$

  $\Rightarrow \bar{x}=\frac{9.9}{9}$

 $\Rightarrow \bar{x}=1.1$

Standard deviation $\sigma$ $=\sqrt{\frac{\sum (x-\bar{x})^2}{n-1}}$

   $\Rightarrow \sigma=\sqrt{\frac{\sum (x-\bar{x})^2}{n-1}}=0.122$

here, $\alpha=1-\;\text{confidence level}$

     $\Rightarrow \alpha=1-0.95=0.05$

$Z_{\frac{\alpha}{2}}=Z_{0.05}=2.306$ by the Z- table

$95\%$ confidence interval is :

    $\bar{x}\pm Z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n}}$

$\Rightarrow 1.1\pm 2.306\times \frac{0.122}{\sqrt{9}}$

$\Rightarrow 1.1\pm 2.306\times \frac{0.122}{3}$

$\Rightarrow 1.1\pm 0.094$

$\Rightarrow (1.006, 1.194)$

Therefore, the mean of the potato chips lie between the interval $(1.006, 1.194)$