# Mean, Standard Deviation, Confidence Intervals

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Graduating students in Business Management at many universities are required to take a the MFAT exam. Below is a random sample of results from five sections of the exam.

Grades

80

90

91

62

77

a. Compute the mean and the standard deviation of the sample.

b. Compute the margin of error at 95% confidence.

c. Develop a 95% confidence interval estimate for the mean of the population. Assume the population is normally distributed.

The American Medical News reports that there are few doctors in rural parts of America. A random sample of 87 rural doctors reveals that on average they make $99,400, with a standard deviation of $12,000.

a. Using the 95% confidence level, what is the value of t-statistic?

b. Develop a 95% confidence interval for the average yearly income of the rural physicians.

c. Construct an 80% confidence interval for the average yearly income for rural physicians.

d. Discuss Why the 95% and 80% intervals are different.

Because of braking problems, Toyota this year recalled the Prius and Lexus cars made between. The models recalled are those that they were made between March and October 2009. Assume that one major dealership sold1200 cars, out of which 18 cars had braking problems and were subsequently recalled.

a. Determine a 95% confidence interval for the percentage of defective cars.

b. If 1.5 million Prius and Lexus were sold between December 2012 and March 2013, determine an interval for the number of defective cars in sales.

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The solution gives detailed steps on solving various questions on confidence intervals. All the formula and calcuations are shown and explained.

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1.

a. Compute the mean and the standard deviation of the sample.

answer: mean=(80+90+91+62+77)/5=80. sd=sqrt{[(80-80)^2+(90-80)^2+(91-80)^2+(62-80)^2+(77-80)^2]/(5-1)]=11.77

b. Compute the margin of error at 95% confidence.

answer: from t table with df=5-1=5 and at 95% level, t value=2.776. hence margin of error=2.776*11.77/sqrt(5)=14.61

c. Develop a 95% confidence interval estimate for the mean of the population. Assume the population is normally distributed.

answer: a 95% confidence interval for mean of the ...

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