1) A food inspector, examining 5 jars of a certain brand of peanut butter, obtained the following percentages of impurities:
2.3; 1.9; 2.1; 2.8; 2.3
Assuming that such determinations are normally distributed, construct a 99% confidence interval for the average percentage of impurities in this brand of peanut butter.

2) Twelve randomly selected citrus trees of one variety have a mean height of 13.8 ft with a standard deviation of 1.2 feet, and fifteen randomly selected citrus trees of another variety have a mean height of 12.9 ft with a standard deviation of 1.5 ft. Stating appropriate assumptions, construct a 95% confidence interval for the difference between the true average heights of the two kinds of citrus trees.

3) A box contains 500 ball bearings which have a mean weight of 5.02 ounces and a standard deviation of 0.3 ounces. A random sample of 100 ball bearings is drawn with replacement from that box. Let x- be the mean weight of the sample of 100 ball bearings. Estimate the probability that x- will be between 4.96 and 5.00 ounces.

4) In a random sample of 120 cheerleaders, 54 suffered moderate to severe damage to their voices. Find a 90% confidence interval for the true proportion of cheerleaders who are afflicted in this way.

5)Miss Prim's class of grade 1 students has 22 children whose mean height is 47.75 inches while Mr. Trim's class of grade 2 students has 25 children whose mean height is 50.4 inches. The standard deviations of the heights of first graders and second graders are known in general to be 1.8 inches and 2.05 inches, respectively. Stating appropriate assumptions, find a 98% confidence interval for the mean growth µ2 -µ1 between second graders and first graders.

Solution Summary

This solution gives the step by step method for computing confidence interval

Which of the following is not needed to be known to calculate a confidenceinterval?
a. standard deviation
b. sample size
c. mean
d. degree of confidence

Compute a 95% confidenceintervalfor the population mean, based on the sample 1.5, 1.54, 1.55, 0.09, 0.08, 1.55, 0.07, 0.99, 0.98, 1.12, 1.13, 1.00, 1.56, and 1.53. Change the last number from 1.53 to 50 and recalculate to the confidenceinterval. Using the results, describe the effect of an outlier or extreme value on the conf

The width of a confidenceinterval estimate for a proportion will be:
A. narrower for 99% confidence than for 95% confidence
B. wider for a sample size of 100 than for a sample size of 50
C. narrower for 90% confidence than for 95% confidence
D. narrower when the sample proportion is 0.50 than when the sample proportion is

Compute a 95% confidenceintervalfor the population mean, based on the sample 25, 27, 23, 24, 25, 24 and 59. Change the number from 59 to 24 and recalculate the confidenceinterval. Using the results, describe the effect of an outlier or extreme value on the confidenceinterval.

Computing from a large sample, one finds the 95% confidenceintervalfor μ to be (6.8, 14.2). Based on this information alone, determine the 90% confidenceintervalfor μ.

A sample of n=16 scores is obtained from an unknown population. The sample has a mean of M=46 with SS=6000.
a. Use the sample data to make an 80% confidenceinterval estimate
of the unknown population mean.
b. Make a 90% confidenceinterval estimate of μ.
c. Make a 95% confidenceinterval estimate of μ.

Assume that in a hypothesis test with null hypothesis H 0: mu = 14.0 at alpha = 0.05, that a value of 13.0 for the sample mean results in the null hypothesis being rejected. That corresponds to a confidenceinterval result of:
a) the 95% confidenceintervalfor the mean contains the value 14.0
b) the 95% confidenceinterval

If x bar =75, S = 24, and n = 36, and assuming that the population is normally distributed, construct a 95% confidenceinterval estimate of the population mean, µ.

Suppose that, for a sample size n = -100 measurements, we find that x = 50. Assuming that the standard deviation equals 2, calculate confidenceintervals for the population mean with the following confidence levels:
a) 95% b) 99% c) 97% d) 80% e) 99.73% f) 92%