Question

In: Statistics and Probability

A recent study examined hearing loss data for 1,771 U.S. teenagers. In this sample, 333 were...

  1. A recent study examined hearing loss data for 1,771 U.S. teenagers. In this sample, 333 were found to have some level of hearing loss. Your goal is to estimate the true long run proportion of all U.S teenagers who have some level of hearing loss.

Suppose we are interested in whether or 1 in 5 teens have hearing loss or not.

  1. In symbols, state the correct hypotheses to test this claim.

  1. Based on your confidence interval above, would you reject or fail to reject the null hypotheses? Why?

  1. What is the average time of a four-man Olympic Bobsleigh team? In the 2014 Sochi Olympics, the 27 finalist teams averaged 3.41 minutes with a standard deviation of 0.42.

  1. Find the observed statistic (also called the point estimate).

  1. Find the margin of error associated with the 95% confidence interval. (3 pts)

  1. Construct a 95% confidence interval for the true long run average time of a four-man Olympic Bobsleigh team.

  1. Interpret the confidence interval in context.

Solutions

Expert Solution

1)

a)

Ho :   p =    0.2
H1 :   p ╪   0.2

b)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   333          
Sample Size,   n =    1771          
                  
Sample Proportion ,    p̂ = x/n =    0.1880          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.009285          
margin of error , E = Z*SE =    1.960   *   0.00928   =   0.0182
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.18803   -   0.01820   =   0.1698
Interval Upper Limit = p̂ + E =   0.18803   +   0.01820   =   0.2062
                  
95%   confidence interval is (   0.170   < p <    0.206   )

CI contain 0.2 , so donot reject Ho

..................

2)

a)

the point estimate =3.41

b)

sample std dev ,    s =    0.4200
Sample Size ,   n =    27

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   26          
't value='   tα/2=   2.056   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   0.42/√27=   0.0808          
margin of error , E=t*SE =   2.0555   *   0.0808   =   0.166

...........

c)

confidence interval is                   
Interval Lower Limit = x̅ - E =    3.41   -   0.1661   =   3.2439
Interval Upper Limit = x̅ + E =    3.41   -   0.1661   =   3.5761
95%   confidence interval is (   3.24   < µ <   3.58   )

d)


There is    95%   confidence that true long run average time of a four-man Olympic Bobsleigh team   lies within confidence interval          

...............

Please let me know in case of any doubt.

Thanks in advance!


Please upvote!

orchestra answered 1 year ago
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