SOLUTION: 5.7 Suppose that the population standard deviation (sigma) for a normally distributed standardized test of achievement is known to be 7.20. You can use these to estimate the sta
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-> SOLUTION: 5.7 Suppose that the population standard deviation (sigma) for a normally distributed standardized test of achievement is known to be 7.20. You can use these to estimate the sta
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Question 1047800: 5.7 Suppose that the population standard deviation (sigma) for a normally distributed standardized test of achievement is known to be 7.20. You can use these to estimate the standard error of the mean for a randomly drawn sample size of 16.
6 5 6 12 5 10 11 13
12 10 9 20 23 28 20 18
But now suppose that we do not feel comfortable assuming that s = 7.20. Use the scores above to
a. estimate the standard error of the sample mean ()
b. find the 95% confidence interval for the mean.
You can put this solution on YOUR website! Suppose that the population standard deviation (sigma) for a normally distributed standardized test of achievement is known to be 7.20. You can use these to estimate the standard error of the mean for a randomly drawn sample size of 16.
6 5 6 12 5 10 11 13
12 10 9 20 23 28 20 18
But now suppose that we do not feel comfortable assuming that s = 7.20. Use the scores above to
a. estimate the standard error of the sample mean ()
s = 6.7
b. find the 95% confidence interval for the mean.
x-bar = 13
ME = 1.96*13/sqrt(16) = 7.37
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95% CI:: 13-7.37 < u < 13+7.37
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c. find the 99% confidence interval for the mean
x-bar = 13
ME = 2.5758*13/sqrt(16) = 8.37
---
99% CI: 13-8.37 < u < 13+8.37
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Cheers,
Stan H.
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You can put this solution on YOUR website! sample mean = 13
sample SD = 6.9 (Generally rounded to one more decimal point than Data values have)
Recommend following to check work on sample means and 'corrected' standard deviations.
www.numberempire.com/statisticscalculator.php
95% confidence interval
Small Sample - We are using the T distribution because
a) population standard deviation is not being used
b) n = 16 < 30
use t-value = 2.13
SE = (2.13)(6.9/sqrt(16)) = (2.13)1.725
Round the confidence interval limits to the same number of decimal places as the sample standard deviation
CI = 13 ± SE
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99% confidence interval
Small Sample
use t-value = 2.95
SE = (2.95)(6.9/sqrt(16)) = (2.95)1.725 = 5.1
Round the confidence interval limits to the same number of decimal places as the sample standard deviation
CI = 13 ± 5.1
(