A random sample of 161 6t\" graders estimated a mean backpack weight of 13.83 lb

Question

A random sample of 161 6t" graders estimated a mean backpack weight of 13.83 lbs. The population standard deviation in backpack weights is assumed to be 6 lbs 1, what is a 90% confidence interval for the mean backpack weight for these data? a) Find the critical z-score b) Calculate the margin of error c) Construct the confidence interval 2, what is a 95% confidence interval for the mean backpack weight for these data? a) Find the critical z-score b) Calculate the margin of error c) Construct the confidence interval 3, What is a 99% confidence interval for the mean backpack weight for these data? a) Find the critical z-score b) Calculate the margin of error c) Construct the confidence interval 4. Which of your three confidence intervals is mostlikely to MISS the true value for the population mean? Why

Explanation / Answer

From the information

Xbar = 13.83 , sigma =6 and n= 161

Question 1.

The 90% confidence interval for the mean backpack weight is

( xbar -Zaplha/2*sigma/sqrt(n) , xbar + Zaplha/2*sigma/sqrt(n))

where Zalpha/2 = critical z-score

and Zalpha/2*sigma/sqrt(n) = margin of error

alpha = level of significance = 0.10

i) Critical Z-score = Zalpha/2 = Z 0.05 = 1.64

ii) Margin of error = Zalpha/2*sigma/sqrt(n) = 1.64 * 6 /sqrt(161) = 0.7751

iii) 90% Confidence interval for population mean

( xbar -Zaplha/2*sigma/sqrt(n) , xbar + Zaplha/2*sigma/sqrt(n)) = ( 13.83 - 0.7751, 13.83 +0.7751)

= ( 13.0545, 14.6055)

For 90 % confidence interval

P( Miss the true value of population mean lies in the interval ) =0.10

Question 2 :

The 95% confidence interval for the mean backpack weight is

( xbar -Zaplha/2*sigma/sqrt(n) , xbar + Zaplha/2*sigma/sqrt(n))

where Zalpha/2 = critical z-score

and Zalpha/2*sigma/sqrt(n) = margin of error

alpha = level of significance = 0.05

i) Critical Z-score = Zalpha/2 = Z 0.025 = 1.96

ii) Margin of error = Zalpha/2*sigma/sqrt(n) = 1.96 * 6 /sqrt(161) = 0.9268

iii) 95% Confidence interval for population mean

( xbar -Zaplha/2*sigma/sqrt(n) , xbar + Zaplha/2*sigma/sqrt(n)) = ( 13.83 - 0.9268, 13.83 + 0.9268)

= ( 12.9032 , 14.7568)

For 95 % confidence interval

P( Miss the true value of population mean lies in the interval ) =0.05

Question 3 :

The 99% confidence interval for the mean backpack weight is

( xbar -Zaplha/2*sigma/sqrt(n) , xbar + Zaplha/2*sigma/sqrt(n))

where Zalpha/2 = critical z-score

and Zalpha/2*sigma/sqrt(n) = margin of error

alpha = level of significance = 0.01

i) Critical Z-score = Zalpha/2 = Z 0.005 = 2.58

ii) Margin of error = Zalpha/2*sigma/sqrt(n) = 2.58 * 6 /sqrt(161) = 1.2199

iii) 95% Confidence interval for population mean

( xbar -Zaplha/2*sigma/sqrt(n) , xbar + Zaplha/2*sigma/sqrt(n)) = ( 13.83 - 1.2199, 13.83 + 1.2199)

= ( 12.6101 , 15.0499)

For 99 % confidence interval

P( Miss the true value of population mean lies in the interval) =0.01

Question 4 :

For 90% confidence interval

P( Miss the true value of population mean lies in the interval ) =0.10 is the maximum

Hence 90% confidence interval is most likely to to miss the true value of population mean.

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