IlT-Mobile Wi-Fi11:12 PM Close WeBWorK : EBF 472. (10 points) A statistics pract

Question

IlT-Mobile Wi-Fi11:12 PM Close WeBWorK : EBF 472. (10 points) A statistics practitioner took a random sample of 54 observations from a population whose standard deviation is 30 and computed the sample mean to be 108. Note: For each confidence interval, enter your answer in the form (LCL UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 95% confidence, changing the population standard deviation to 45; Confidence Interval C. Estimate the population mean with 95% confidence, changing the population standard deviation to 7; Confidence Interval Note: You can earn partial credit on this

Explanation / Answer

From the information

Xbar = 108 and n= 54 , sigma = 30

A) 95% Confidence interval for population mean is

( xbar - ( sigma/ sqrt(n) * Z 0.025) , xbar +( sigma/ sqrt(n) * Z 0.025) )

From normal table Z 0.025 = 1.96

( 108 - ( 30 /sqrt(54) * 1.96) , 108+ ( 30 /sqrt(54) * 1.96))

= ( 99.9983 , 116.0017)

Confidence Interval = ( 99.9983, 116.0017)

B) If Standard deviation sigma =45

Confidence interval is

( 108 - ( 45 /sqrt(54) * 1.96) , 108 + ( 45 /sqrt(54) * 1.96))

= (95.9975, 120.0025)

Confidence Interval = (95.9975, 120.0025)

C) If Standard deviation sigma =7

Confidence interval is

( 108 - ( 7 /sqrt(54) * 1.96) , 108 + ( 7 /sqrt(54) * 1.96))

= (106.1329, 109.8671)

Confidence Interval = (95.9975, 120.0025)

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