Answer :
Answer:
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 =
[tex]s^2=0.44[/tex]
Significance level = 0.02
Degree of freedom = n - 1 = 53
We use chi-square to find the confidence interval for population variance.
[tex]\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[/tex]
Formula:
[tex]\dfrac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2}}}< \sigma^2 < \dfrac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2}}}[/tex]
Putting values, we get,
[tex]\dfrac{53\times 0.44}{79.843}<\sigma^2< \dfrac{53\times 44}{32.018}\\\\0.29207<\sigma^2 < 0.72834\\\approx 0.29<\sigma^2< 0.73[/tex]
is the required 98% confidence interval for the variance of the weights of the packages prepared by the machine.