1. x̄ = 85 and σ = 8, and n = 64, set up a 95% confidence interval estimate of the population mean μ.
85-1.96(8/64)< μ < 85 + 1.96(8/64)
83.04 < μ < 86.96
2. If x̄ = 125, σ = 24 and n = 36, set up a 99% confidence interval estimate of the population mean μ.
125-2.58(24/36)< μ < 125 + 2.58(24/36)
114.68 < μ < 135.32
3. The manager of a supply store wants to estimate the actual amount of paint contained in 1-gallon cans purchased from a nationally known manufacturer. It is known from the manufacturer's specification sheet that standard deviation of the amount of paint is equal to 0.02 gallon. A Random sample of 50 cans is selected and the sample mean amount of paint per 1 gallon is 0.99 gallon.
3a. Set up a 99% confidence interval estimate of the true population mean amount of paint included in 1-gallon can?
3b. On the basis of your results, do you think that the manager has a right to complain to the manufacturer? why?
The paint in the gallons is being filled up to the .02 standard deviation of the gallon, so the manufacturer is telling the truth and the manager can't complain.
4. A stationery store wants to estimate the mean retail value of greeting cards that has in its inventory. A random sample of 20 greeting cards indicates an average value of $1.67 and standard deviation of $0.32
4a. Assuming a normal distribution set up with 95% confidence interval estimate of the mean value of all greeting cards stored in the store's inventory.
1.96(.32/√20) = .14
1.67 + .14 = 1.81
1,67 - .14 = 1.53
1.53 < μ < 1.81
4b. How might the result obtained in (a) be useful in assisting the store owner to estimate of the mean value of all greeting cards in the store's inventory.
The answer in 4a shows the store owner that 95% of all cards are valued between $1.53 and $1.81.
5. If you want to be 95% confident of estimating the population mean to within a sampling error of ± 5 and standard deviation is assumed to be equal 15, what sample size is required?
(1.96 * 15/5)^2 = 35
sample size = 35
6. Generate your own null and alternative hypothesis statements and provide rationale for your selection.
Alternative: Adults who drink 8 glasses of water per day live longer lives than those who do not.
Null: Adults who drink 8 glasses of water per day DO NOT live longer lives than those who do not.
85-1.96(8/64)< μ < 85 + 1.96(8/64)
83.04 < μ < 86.96
2. If x̄ = 125, σ = 24 and n = 36, set up a 99% confidence interval estimate of the population mean μ.
125-2.58(24/36)< μ < 125 + 2.58(24/36)
114.68 < μ < 135.32
3. The manager of a supply store wants to estimate the actual amount of paint contained in 1-gallon cans purchased from a nationally known manufacturer. It is known from the manufacturer's specification sheet that standard deviation of the amount of paint is equal to 0.02 gallon. A Random sample of 50 cans is selected and the sample mean amount of paint per 1 gallon is 0.99 gallon.
3a. Set up a 99% confidence interval estimate of the true population mean amount of paint included in 1-gallon can?
.99-2.58(.02/√50)< μ < 99+2.58(.02/√50)
.98 < μ < 1
3b. On the basis of your results, do you think that the manager has a right to complain to the manufacturer? why?
The paint in the gallons is being filled up to the .02 standard deviation of the gallon, so the manufacturer is telling the truth and the manager can't complain.
4. A stationery store wants to estimate the mean retail value of greeting cards that has in its inventory. A random sample of 20 greeting cards indicates an average value of $1.67 and standard deviation of $0.32
4a. Assuming a normal distribution set up with 95% confidence interval estimate of the mean value of all greeting cards stored in the store's inventory.
1.96(.32/√20) = .14
1.67 + .14 = 1.81
1,67 - .14 = 1.53
1.53 < μ < 1.81
4b. How might the result obtained in (a) be useful in assisting the store owner to estimate of the mean value of all greeting cards in the store's inventory.
The answer in 4a shows the store owner that 95% of all cards are valued between $1.53 and $1.81.
5. If you want to be 95% confident of estimating the population mean to within a sampling error of ± 5 and standard deviation is assumed to be equal 15, what sample size is required?
(1.96 * 15/5)^2 = 35
sample size = 35
6. Generate your own null and alternative hypothesis statements and provide rationale for your selection.
Alternative: Adults who drink 8 glasses of water per day live longer lives than those who do not.
Null: Adults who drink 8 glasses of water per day DO NOT live longer lives than those who do not.
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