Confidence level and margin of error. Example 14.8 (textbook page 350) described

Question

Confidence level and margin of error. Example 14.8 (textbook page 350) described a quality control study of the aspirin content of aspirin tablets. We treated the aspirin content of tablets as normally distributed with standard deviation = 5 mg.

a). Give a 95% confidence interval for the mean aspirin content in this population. The actual sample, a simple random sample of 10 tablets, gave x = 326.9 mg.

b). Now give the 90% and 99% confidence intervals for
What are the margins of error for 90%, 95% and 99% confidence? How does increasing the confidence level change the margin of error of a confidence interval?

c) The method you use for a) and b) is _________(Z or T) method? Explain.

Explanation / Answer

a)

b)

90% confidence interval

99% confidence interval

Increasing the confidence inteval results into increasing the value of margin of error.

c)

Here t method is used as n<30

CI for 95% n 10 mean 326.9 t-value of 95% CI 2.2622 std. dev. 5 SE = std.dev./sqrt(n) 1.58114 ME = t*SE 3.57678 Lower Limit = Mean - ME 323.32322 Upper Limit = Mean + ME 330.47678 95% CI (323.3232 , 330.4768 )

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