Calculate the sample mean and sample standard deviation.

Question

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Here are the solutions to the questions:

a) Calculate a 96% confidence interval for the mean weight of all cabbages.

Calculate the sample mean and sample standard deviation.
Given:
Sample size, $n = 100$
Sum of weights, $\sum x = 16380$
Sum of squared weights, $\sum x^2 = 2,686,310$

Sample mean:
$\bar{x} = \frac{\sum x}{n} = \frac{16380}{100} = 163.8$
Sample variance:
$s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$ $s^2 = \frac{1}{100-1} \left( 2,686,310 - \frac{(16380)^2}{100} \right)$ $s^2 = \frac{1}{99} \left( 2,686,310 - \frac{268304400}{100} \right)$ $s^2 = \frac{1}{99} \left( 2,686,310 - 2,683,044 \right)$ $s^2 = \frac{3266}{99} \approx 33.00$
Sample standard deviation:
$s = \sqrt{33.00} \approx 5.7446$
Determine the critical z-value.
For a 96% confidence level, $\alpha = 1 - 0.96 = 0.04$ .
The critical value is $z_{\alpha/2} = z_{0.02}$ .
Looking up the z-table for an area of $1 - 0.02 = 0.98$ , we find $z_{0.02} \approx 2.05$ .
Calculate the margin of error.
The margin of error ( $E$ ) is given by:
$E = z_{\alpha/2} \frac{s}{\sqrt{n}}$ $E = 2.05 \times \frac{5.7446}{\sqrt{100}}$ $E = 2.05 \times \frac{5.7446}{10}$ $E = 2.05 \times 0.57446 \approx 1.1776$
Construct the confidence interval.
The confidence interval is $\bar{x} \pm E$ :
$163.8 \pm 1.1776$
Lower bound: $163.8 - 1.1776 = 162.6224$
Upper bound:

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