Question 1097914: In a random sample of four microwave ovens, the mean repair cost was $80.0080.00 and the standard deviation was $11.50. Assume the population is normally distributed and use a t-distribution to construct a 95% confidence interval for the population mean μ. What is the margin of error of μ? Interpret the results.
The 95% confidence interval for the population mean muμ is ( ).
The margin of error is
(Round to two decimal places as needed.)
Interpret the results. Choose the correct answer below.
A.With 95% confidence, it can be said that the population mean repair cost is between the bounds of the confidence interval.
B.With 95% confidence, it can be said that the repair cost is between the bounds of the confidence interval.
C.It can be said that 95% of microwaves have a repair cost between the bounds of the confidence interval.
D.If a large sample of microwaves are taken approximately 95% of them will have repair costs between the bounds of the confidence interval.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! the sample mean of the 4 microwaves is $80 with a standard deviation of $11.50
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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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standard error(SE) = sample standard deviation / square root(sample size)
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SE = 11.50 / square root(4) = 5.75
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degrees of freedom(DF) = 4 - 1 = 3
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critical value(CV) = 3.182
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Note CV is the t-statistic for the DF and p*
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margin of error(ME) = CV * SE = 3.182 * 5.75 = 18.2965 approximately 18.30
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95% confidence interval is $80 + or - 18.30, the interval is ($61.70, $98.30)
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a 95% confidence interval means that if we take samples and compute their 95% confidence intervals, then we expect that some interval estimates would include the true population mean and some would not. Therefore, for this problem we expect 95% of the computed intervals to include the population mean.
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The correct answer for interpretation of a 95% confidence interval is D.
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