Question 1149929: Waiting times (in hours) at a particular hospital are believed to be approximately normally distributed with a variance of 2.25 hr.
a.. A sample of 20 outpatients revealed a mean waiting time of 1.52 hours. Construct the 95% CI for the estimate of the population mean.
b. Suppose that the mean of 1.52 hours had resulted from a sample of 32 patients. Find the 95% CI.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! a. sample mean(u) of 1.52
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standard error(SE) of the sample = 2.25/square root(20) = 0.5031
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alpha(a) = 1 -(95/100) = 0.05
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critical probability(p*) = 1 -a/2 = 0.975
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degrees of freedom(DF) = 20 -1 = 19
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because the sample size(20) is less than 30, we use the student t-distribution tables to find the critical value(CV) which is the t-score for DF=19 and p* = 0.975. CV = 2.093
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Margin of Error(ME) = CV * SE = 2.093 * 0.5031 = 1.053
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For a normal distribution the population mean is also the mean of the sample
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95% confidence interval(CI) = 1.52 + or minus 1.053 = (0.467, 2.573)
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b. SE = 2.25/square root(32) = 0.3977
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p* = 0.975
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because the sample size(32) is greater than 30, we can use the Normal distribution table of z-values to find the CV which is the z-score associated with the p*. CV = 1.96
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ME = 1.96 * 0.3977 = 0.7795
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95% confidence interval(CI) = 1.52 + or minus 0.7795 = (0.7405, 2.2995)
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