Please help with this homework question 6.4.20-1Question HelpAs part of a survey, a marketing representative asks a random sample of 29 business owners how much they would be willing to pay for a website for their company. Shefinds that the sample standard deviation is $3376. Assume the sample is taken from a normally distributed population. Construct 90% confidence intervals for (a) thepopulation variance o2 and (b) the population standard deviation o. Interpret the results.(a) The confidence interval for the population variance is (, ).(Round to the nearest integer as needed.)

Name: Please help with this homework question 6.4.20-1Question HelpAs part o
Uploaded: 2024-03-20T07:07:37.000Z
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Description: Please help with this homework question 6.4.20-1Question HelpAs part of a survey, a marketing representative asks a random sample of 29 business owners how much they would be willing to pay for a website for their company. Shefinds that the sample st

The provided values are: Sample size n=29 Sample standard deviation s=$3376 Degrees of freedom: The degrees of freedom are calculated as given below: n-1=28 Confidence level =0.90 Level of significance is 0.10 Critical value: Using chi-square table, the critical values are as follows: χleft2=χ0.05,282=41.337 χright2=χ0.95,282=16.928 Calculation: (a) 90% confidence interval for population variance: The 90% confidence interval for population variance can be calculated as follows: n-1s2χleft2<σ2<n-1s2χright229-13376241.337<σ2<29-13376216.9287720118<σ2<18851992 Interpretation: There is 90% confidence that the population variance lie between 7720118 and 18851992.(b) 90% confidence interval for population standard deviation: The 90% confidence interval for population standard deviation can be calculated as follows: n-1s2χleft2<σ<n-1s2χright229-13376241.337<σ<29-13376216.9287720118.25<σ<18851992.442779<σ<4342 Interpretation: There is 90% confidence that the population standard deviation lie between 2779 and 4342.