Question

Based on a random sample of 20 students from a PE class, they exercise 1.6 hours (96 minutes) per day on average with sample standard deviation 1.23 hours (78 minutes).

Use this information to construct a 95% confidence interval to estimate  μ, the population mean number of hours of daily exercise for students.

a. State the assumption(s)

b. identify whether each assumption is met or not.

c. Show your work to calculate the confidence interval and

d. interpret the result in the context of the problem

Answer 1

a)

Population must be roughly symmetric--sample<30              
  

b)

Assumption is met

c)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   19          
't value='   tα/2=   2.093   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   1.23/√20=   0.2750          
margin of error , E=t*SE =   2.0930   *   0.2750   =   0.576
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    1.60   -   0.5757   =   1.0243
Interval Upper Limit = x̅ + E =    1.60   -   0.5757   =   2.1757
95%   confidence interval is (   1.0   < µ <   2.2   )

d)

There is    95%   confidence that true mean lies within confidence interval              

Please let me know in case of any doubt.

Thanks in advance!

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