Mathematics: A frozen food company uses a machine that packages okra in six ounce portions. A sample of 5454 packages of okra has a variance of 0.440.44. Construct the 98%98% confidence interval to estimate the variance of the weights of the packages prepared by the machine. Round your answers to two decimal places.

If 5 for every 7 (ratio 5:7)Therefore 56/7=8So 8 red marbles and 56 blueI hope this helps...

A frozen food company uses a machine that packages

bakoeboo

22.01.2022

98% confidence interval: (0.29, 0.73)

Step-by-step explanation:

We are given the following in the question:

Sample size,n = 54

Sample variance =

s^2=0.44

Significance level = 0.02

Degree of freedom = n - 1 = 53

We use chi-square to find the confidence interval for population variance.

$\chi^2_{\frac{\alpha}{2}} = \chi^2_{(0.01,53)} = 79.843\\\\\chi^2_{\frac{1-\alpha}{2}} = \chi^2_{(0.99,53)} =32.018$

Formula:

$\dfrac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2}}}< \sigma^2 < \dfrac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2}}}$

Putting values, we get,

$\dfrac{53\times 0.44}{79.843}$

is the required 98% confidence interval for the variance of the weights of the packages prepared by the machine.

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