Confidence Interval Estimation And introduction to Fundamental of hypothesis testing

1. x̄ = 85 and σ = 8, and n = 64, set up a 95% confidence interval estimate of the population mean μ. 
Z= 1-(0.05/2) = 1.96

Sample mean= x-bar = 85

Z*s/sqrt(n) = (1.96*8)/sqrt(64) = 1.96

CI= 85 – 1.96= 83.04

CI= 85- 1.96= 86.96

(83.04, 86.96)

2. If  x̄ = 125, σ = 24 and n = 36, set up a 99% confidence interval estimate of the population mean μ. 

Z= 1- (0.01/2) = 0.995= 2.57

Z*s/sqrt(n) = 125 - (2.57*8/sqrt(36) = 3.42-125= 121.58

Z*s/sqrt(n) = 125 + (2.57*8/sqrt(36) = 3.42+125= 128.42

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?

Z= (1-.95)= .05

CI= 0.998+ 1.96(0.02/sqrt(50)= 1.0035

CI= 0.998-1.96(0.02/sqrt(50)= 0.9925

(0.99, 1.00)

3b. On the basis of your results, do you think that the manager has a right to complain to the manufacturer? why? 

The 95% confidence interval estimate for the population mean amount of paint included in a 1-gallon cans is (0.99, 1.00). so no the manager has right to complain because paint include 1 gallon falls in 95  interval.

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.

x = 1.67

ME= (1.96)(0.32/sqrt(20)) = 0.1402

CI= 1.67- 01402 = 1.53, 1.67+01402= 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 X cards store value fall between 1.53 and 1.81 dollars.

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? 

E=5

S= 15

ME(1.96*15/5)^2= 34.57