As part of a survey, a marketing representative asks a random sample of 28 busin

Question

As part of a survey, a marketing representative asks a random sample of 28 business owners how much they would be willing to pay for a website for their company. She finds that the sample standard deviation is $3349. Assume the sample is taken from a normally distributed population. Construct 90% confidence intervals for a the population variance and b) the population standard deviation Interpret the results (a) The confidence interval for the population variance is ( Round to the nearest integer as needed.) Interpret the results. Select the correct choice below and fill in the answer box(es) to complete your choice Round to the nearest integer as needed.) d) A. With 90% confidence, you can say that the population variance is between O B. With 90% confidence, you can say that the population variance is greater than and ° C. With 10% confidence, you can say that the population vanance is between D, with 10% confidence, you can say that the population variance is less than and (b) The confidence interval for the population standard deviation is ( Round to the nearest integer as Interpret the results. Select the correct choice below and fill in the answer box(es) to complete your choice Round to the nearest integer as O B. With 10% conndence, you can say that the population standard deviations greater than $ O D with 90% confidence, you can say that the population standard deviation is between S and $ O A. With 90% confidence, you can say that the population standard deviation is less than $ C. With 10% confidence, you can say that the population standard deviation is between S and $L

Explanation / Answer

PART A.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ^2 right < ^2 < (n-1) s^2 / ^2 left
where,
s = standard deviation
^2 right = (1 - confidence level)/2
^2 left = 1 - ^2 right
n = sample size
since alpha =0.1
^2 right = (1 - confidence level)/2 = (1 - 0.9)/2 = 0.1/2 = 0.05
^2 left = 1 - ^2 right = 1 - 0.05 = 0.95
the two critical values ^2 left, ^2 right at 27 df are 40.1133 , 16.151
s.d( s )=3349
sample size(n)=28
confidence interval for ^2= [ 27 * 11215801/40.1133 < ^2 < 27 * 11215801/16.151 ]
= [ 302826627/40.1133 < ^2 < 302826627/16.1514 ]
= [ 7549282.3328 < ^2 < 18749249.4149 ]
~ [ 7549282 < ^2 < 18749249 ]

with 90% confidence you can say that the 90% population
variance is between 7549282 and 18749249

PART B.
confidence interval for = sqrt(lower) < < sqrt(upper)
= [ sqrt (7549282.3328) < < sqrt(18749249.4149), ]
= [ 2747.5957 < < 4330.0403 ]
~ [ 2748< < 4330 ]

with 90% confidence you can say that the 90% population
standard deviation is between 2748 and 4330

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