Question

help with 1-10 please!!!!

You are doing marketing research about yoqurt consumption Based on your previous research you feel sure that the population standard deviation of the number of yogurts consumed per year is 6. In other words, is known and = 6 You take a sample of 50 consumers and the sample mean is 32 yogurts consumed per year in other words. 32. You want to develop a confidence interval for the population mean) such that you are 90% confident that the true value of lies within the 1. What is the value of ? 2. What is the value of 2 3. What is the value of 2..? 1. What is the value of the margin of error (MOE)? 5. State the confidence interval showing your confidence level that lies between two numbers 6. Redo the problem for a 95% confidence interval 7. Redo the problem for a 99% confidence interval 8. What happens to the margin of me and the confidence interval as the confidence level increases? You took over William Gosset's old and you are measuring the quantity of a certain enzyme found in Guiness beer. You are pretty sure that the distribution of this in bones of beer has a normal distribution. You take a small sample of bottles of Guiness and find the following values for the quantity of the enzyme: 10 8 12 15 13 11 6 5 You want to determine a range of values for the mean quantity of this enzyme and you want to be 95% confident that your range contains the true population mean 9. What is your point estimate for the population mean quantity of this enzyme in Guiness? 10. What is your point estimate for the population standard deviation of the quantity of this enzyme in Guiness?

Answer 1

1)α=0.10

2)α/2 = 0.05

3) z value=   z α/2=   1.6449   [Excel formula =NORMSINV(α/2) ]

4) Standard Error , SE = σ/√n =   6.0000   / √   50   =   0.8485
margin of error, E=Z*SE =   1.6449   *   0.8485   =   1.3957

5) confidence interval is                   
Interval Lower Limit = x̅ - E =    32.00   -   1.395705   =   30.6043
Interval Upper Limit = x̅ + E =    32.00   -   1.395705   =   33.3957
90%   confidence interval is (   30.604   < µ <   33.396   )

6) z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   6.0000   / √   50   =   0.8485
margin of error, E=Z*SE =   1.9600   *   0.8485   =   1.6631
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    32.00   -   1.663085   =   30.3369
Interval Upper Limit = x̅ + E =    32.00   -   1.663085   =   33.6631
95%   confidence interval is (   30.337   < µ <   33.663   )

7)

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.5758   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   6.0000   / √   50   =   0.8485
margin of error, E=Z*SE =   2.5758   *   0.8485   =   2.1857
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    32.00   -   2.185664   =   29.8143
Interval Upper Limit = x̅ + E =    32.00   -   2.185664   =   34.1857
99%   confidence interval is (   29.814   < µ <   34.186   )

8)

as confidence level increases, margin of error and confidence interval increases