Question

1. The average time to drive from Joliet to Chicago in non-rush hour traffic on I 55 is 48 minutes with a sample standard deviation of 6.3 minutes. For n = 65, find the 95% confidence interval for the mean.

2. Meijer has 10 lbs. of potatoes on sale for Thanksgiving. If the bags have a population standard deviation of .75 lbs., find the 90% confidence interval for the mean for n = 50.

3. At JJC, 125 out of 200 students take Communications 101 their first semester at JJC. Find the 93% confidence interval for all first year students at JJC taking Communications 101.

4. The average cost of a large pizza at Dominoes is $19 with a sample standard deviation of $1.75 depending on the number of toppings. For n = 100, find the 98% confidence interval for the price of the pizza.

5. In a first grade class, 9 of 24 students drink white milk with their lunch. Find the 92% confidence interval for students drinking milk.

6. In an elementary school, 7 of 100 students have a peanut allergy. Find the 95% confidence interval for peanut allergies in elementary school districts.

7. The Bears have won an average of 7 games per year from 2010 to 2017 with a population standard deviation of 2.7255. For n = 16, find a 96% confidence interval for the number of wins in the 2018 season.

8. The Chicago Cubs have won an average of 83 games per year from 2008 to 2018 with a sample standard deviation of 14.4844. For n = 162, find the 98% confidence interval for the number of wins in the 2019 season.

9. The average number of hurricanes along the Atlantic coast is 6 with a population standard deviation of 2 per year. For n = 20, find the 95% confidence interval for the number of hurricanes striking the Atlantic coast per year.

10. The average snowfall at O’Hare Airport is 37.1 inches with a sample standard deviation of 5.3. For n = 10, find the 90% confidence interval for the amount of snow anticipated for the winter of 2018 to 2019.

Answer 1

1. The average time to drive from Joliet to Chicago in non-rush hour traffic on I 55 is 48 minutes with a sample standard deviation of 6.3 minutes. For n = 65, find the 95% confidence interval for the mean.

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   64          
't value='   tα/2=   1.998   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   6.3/√65=   0.7814          
margin of error , E=t*SE =   1.9977   *   0.7814   =   1.561
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    48.00   -   1.5611   =   46.4389
Interval Upper Limit = x̅ + E =    48.00   -   1.5611   =   49.5611
95%   confidence interval is (   46.4   < µ <   49.6   )

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Meijer has 10 lbs. of potatoes on sale for Thanksgiving. If the bags have a population standard deviation of .75 lbs., find the 90% confidence interval for the mean for n = 50.

Level of Significance ,    α =    0.1          

z value=   z α/2=   1.645   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   0.75/√50=   0.1061          
margin of error, E=Z*SE =   1.6449   *   0.1061   =   0.174
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    10.00   -   0.1745   =   9.8255
Interval Upper Limit = x̅ + E =    10.00   -   0.1745   =   10.1745
90%   confidence interval is (   9.8   < µ <   10.2   )

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At JJC, 125 out of 200 students take Communications 101 their first semester at JJC. Find the 93% confidence interval for all first year students at JJC taking Communications 101.

Level of Significance,   α =    0.07          
Number of Items of Interest,   x =   125          
Sample Size,   n =    200          
                  
Sample Proportion ,    p̂ = x/n =    0.6250          
z -value =   Zα/2 =    1.812   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.034233          
margin of error , E = Z*SE =    1.812   *   0.03423   =   0.0620
                  
93%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.62500   -   0.06203   =   0.5630
Interval Upper Limit = p̂ + E =   0.62500   +   0.06203   =   0.6870
                  
93%   confidence interval is (   0.563   < p <    0.687   )

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The average cost of a large pizza at Dominoes is $19 with a sample standard deviation of $1.75 depending on the number of toppings. For n = 100, find the 98% confidence interval for the price of the pizza.

Level of Significance ,    α =    0.02          
degree of freedom=   DF=n-1=   99          
't value='   tα/2=   2.365   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   1.75/√100=   0.1750          
margin of error , E=t*SE =   2.3646   *   0.1750   =   0.414
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    19.00   -   0.4138   =   18.5862
Interval Upper Limit = x̅ + E =    19.00   -   0.4138   =   19.4138
98%   confidence interval is (   18.6   < µ <   19.4   )

Please let me know in case of any doubt.

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