Question

In: Statistics and Probability

1a. A study of 24 randomly chosen dishwashers in Lincoln finds that the mean repair cost...

1a. A study of 24 randomly chosen dishwashers in Lincoln finds that the mean repair cost is $92.04, with a sample standard deviation of $12.26. Repair costs are known to be normally distributed. Find a 99% confidence interval for the mean repair cost of all dishwashers in Lincoln. Round the confidence interval values to the same number of decimal places as the sample mean.

1b. In a survey of 240 randomly selected high school wrestlers, 50 were injured during their junior year. Find a 90% confidence interval for the proportion of all high school wrestlers injured during their junior year. Round all proportions to 3 decimal places.

2. Choose ONE of the confidence intervals you found in #1. Write a sentence that gives the practical interpretation of that confidence interval.

3. A quality control inspector claims that over 10% of microwaves need repairs during the first five years of use. A researcher is testing this claim.

a. Write the null and alternative hypotheses, and tell whether the test is left-tailed, right-tailed, or two-tailed.

b. In a random sample of 114 microwaves, 16 needed repairs during the first five years. Find the test statistic.

c. Find the P-value.

d. The level of significance for this hypothesis test is 0.05. Should the researcher reject the null hypothesis or fail to reject the null hypothesis?

e. Write a conclusion in the context of the quality control inspector’s claim about microwaves.

Solutions

Expert Solution

Q 1 ) A) Given that

we want to construct the 99% C.I for mean repair cost of all dishwashers in Licoln

Given that

Sample mean

Sample SD

Sample size = n = 24

Confidence level = C = 0.99

95% C.I for mean is

Where ,

DF = n - 1 = 24 - 1 = 23

tc = 2.807  ..............( From t table )

95% C.I for mean is

Therefore 99% C.I for mean repair cost of all dishwashers in Licoln is ..............(Answer)

======================================================================================

Q1) b)

Given that

sample size = n = 240

Confidence level = C = 0.90

Let x=Number of high school wrestlers injured during their junior year = 50

Thereofore sample proportion is

The 90% confidence interval for the proportion of all high school wrestlers injured during their junior year is

Where

Zc= Critical value of C=1.645 ...............( By using Z table)

we get

The 90% confidence interval for the proportion of all high school wrestlers injured during their junior year is

..........(Answer)

======================================================================================

Q 2) Interpretation for 1b): If we repeatedly perform the experiment large number of times and confidence interval is calculated for each sample then at 90% Confidence level the true proportion of all high school wrestlers injured during their junior year is lies between ..........(Answer)


Related Solutions

A manager records the repair cost for 4 randomly selected TVs. A sample mean of $88.46...
A manager records the repair cost for 4 randomly selected TVs. A sample mean of $88.46 and standard deviation of $17.20 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
A researcher records the repair cost for 10 10 randomly selected VCRs. A sample mean of...
A researcher records the repair cost for 10 10 randomly selected VCRs. A sample mean of $79.18 $ ⁢ 79.18 and standard deviation of $11.06 $ ⁢ 11.06 are subsequently computed. Determine the 90% 90 % confidence interval for the mean repair cost for the VCRs. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Step 2 of 2...
A manager records the repair cost for 4 randomly selected TVs. A sample mean of $88.46...
A manager records the repair cost for 4 randomly selected TVs. A sample mean of $88.46 and standard deviation of $17.20 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal. Step 2 of 2: Construct the 99% confidence interval. Round your answer to two decimal places.
A manager records the repair cost for 29 randomly selected washers. A sample mean of $84.87...
A manager records the repair cost for 29 randomly selected washers. A sample mean of $84.87 and sample standard deviation of $12.34 are subsequently computed. Assume the population is normally distributed. Determine the upper endpoint of a 95% confidence interval estimate of the true mean repair cost. A)84.87 B)88.83 C)86.83 D)30.89 E)89.56
A researcher records the repair cost for 27 randomly selected refrigerators. A sample mean of $97.89...
A researcher records the repair cost for 27 randomly selected refrigerators. A sample mean of $97.89 and standard deviation of $17.53 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
A manager records the repair cost for 29 randomly selected washers. A sample mean of $84.87...
A manager records the repair cost for 29 randomly selected washers. A sample mean of $84.87 and sample standard deviation of $12.34 are subsequently computed. Assume the population is normally distributed. Determine the upper endpoint of a 95% confidence interval estimate of the true mean repair cost. A. 86.83 B. 84.87 C. 89.56 D. 30.89 E. 88.83
A student records the repair cost for 39 randomly selected TVs. A sample mean of $52.20...
A student records the repair cost for 39 randomly selected TVs. A sample mean of $52.20 and standard deviation of $22.66 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the TVs. Assume the population is normally distributed. Step 2 of 2 : Construct the 90% confidence interval. Round your answer to two decimal places.
A student records the repair cost for 13 randomly selected TVs. A sample mean of $72.19...
A student records the repair cost for 13 randomly selected TVs. A sample mean of $72.19 and standard deviation of $15.88 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal. 1) Find the critical value that should be used in constructing the confidence interval. Round you answer to three decimal places. 2) Construct the 98% confidence interval. Round your answer to two decimal places.
A consumer affairs investigator records the repair cost for 4 randomly selected TVs. A sample mean...
A consumer affairs investigator records the repair cost for 4 randomly selected TVs. A sample mean of $75.89 and standard deviation of $13.53 are subsequently computed. Determine the 80% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Step 2 of 2: Construct the 80%80% confidence interval. Round your...
If, for a sample in which the subjects are randomly chosen, the mean income is $45,000,...
If, for a sample in which the subjects are randomly chosen, the mean income is $45,000, the sample size is 1600, the standard deviation is $4,000, and the standard error of the mean is $______, what, approximately, is the 95% confidence interval for the population mean? (show all work) (a) $41,000 to $49,000. (b) $37,000 to $53,000. (c) $44,800 to $45,200. (d) It is impossible to know, because we do not know if the incomes are normally distributed.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT