Question

Can Anyone help me with these 7 questions?

Questions: Q1. A random sample has been taken from a normal distribution and the following confidence intervals constructed using the same data: (38.02, 61.98) and (39.95, 60.05) (a) What is the value of the sample mean? (b) One of these intervals is a 95% CI and the other is a 90% CI. Which one is the 95% CI? Why? Q2. A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with 62 = 1000(psi). A random sample of 12 specimens has a mean compressive strength of x - 3250 psi. (a) Construct a 95% two-sided confidence interval on mean compressive strength. (b) Construct a 99% two-sided confidence interval on mean compressive strength. Compare the width of this confidence interval with the width of the one found in part (a). Q3. A research engineer for a tire manufacturer is investigating tire life for a new rubber compound and has built 16 tires and tested them to end-of-life in a road test. The sample mean and standard deviation are 60,139.7 and 3645.94 kilometers. Find a 95% confidence interval on mean tire life. Q4. The percentage of titanium in an alloy used in aerospace castings is measured in 51 domly selected parts. The sample standard deviat The sample standard deviation is s = 0.37. Construct a 95% two-sided confidence interval for o. Q5. The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 13 defectives (a) Calculate a 95% two-sided CI on the fraction of defective circuits produced by this particular tool. (b) Calculate a 95% upper confidence bound on the fraction of defective circuits. Q6. Of 1000 randomly selected cases of lung cancer, 823 resulted in death within 10 years. (a) Calculate a 95% two-sided confidence interval on the death rate from lung cancer (b) Using the point estimate of p obtained from the preliminary sample, what sample size is needed to be 95% confident that the error in estimating the true value of p is les (c) How large must the sample be if you wish to be at least 95% confident that the error in estimating p is less than 0.03, regardless of the true value of p? Q7. Consider the tire-testing data described in Q3. Compute a 95% prediction interval on the life of the next tire of this type tested under conditions that are similar to those employed in the original test. Compare the length of the prediction interval with the length of the 95% CI on the population mean.

Answer 1

1)

a)

confidence interval is                  
lower limit =    38.02              
upper limit=   61.98              
sample mean = (lower limit+upper limit)/2= (   38.02   +   61.98   ) / 2 =   50

b)

95% CI is wider than 90% CI

so, 95% CI is (38.02 , 61.98)

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2)

a)

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.960   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   31.6228   / √   12   =   9.1287
margin of error, E=Z*SE =   1.9600   *   9.1287   =   17.8919
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    3250.00   -   17.891941   =   3232.1081
Interval Upper Limit = x̅ + E =    3250.00   -   17.891941   =   3267.8919
95%   confidence interval is (   3232.11   < µ <   3267.89   )

b)

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.576   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   31.6228   / √   12   =   9.1287
margin of error, E=Z*SE =   2.5758   *   9.1287   =   23.5140
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    3250.00   -   23.513997   =   3226.4860
Interval Upper Limit = x̅ + E =    3250.00   -   23.513997   =   3273.5140
99%   confidence interval is (   3226.49   < µ <   3273.51   )

99% CI has greater width than 95% CI

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

3)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   15          
't value='   tα/2=   2.131   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   3645.9400   / √   16   =   911.4850
margin of error , E=t*SE =   2.1314   *   911.4850   =   1942.7843
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    60139.70   -   1942.784289   =   58196.9157
Interval Upper Limit = x̅ + E =    60139.70   -   1942.784289   =   62082.4843
95%   confidence interval is (   58196.92   < µ <   62082.48   )