Question

This is an 8 Part Question. If you guys want to count it as 8 question on my chegg account please do so. I know it is long I just dont understand it

The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of females in the labor force is the same in the United States. Representatives from the United States Department of Labor plan to check a random sample of over 10,000 employment records on file to estimate a percentage of females in the United States labor force.

a) They select a random sample of 525 employment records and find that 229 of the people are females. Check the required conditions to construct a confidence interval.

b) Create a 90% confidence interval.

c) Interpret the confidence interval in this context.

d) Explain what 90% confidence means in this context.

e) Should the representatives from the Department of Labor conclude that the percentage of females in their labor force is lower than Europe’s rate of 46%? Explain.

Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. The insurance company knows that, last year, the life expectancy of its policyholders was 77 years. They want to know if their clients this year have a longer life expectancy, on average, so the company randomly samples some of the recently paid policies to see if the mean life expectancy of policyholders has increased. The insurance company will only change their premium structure if there is evidence that people who buy their policies are living longer than before.

86	75	83	84	81	77	78	79	79	81
76	85	70	76	79	81	73	74	73	83

f) To begin to answer this question, we will use a confidence interval. Find and interpret a 95% confidence interval for the mean life expectancy of these policyholders. Make sure to check the conditions required for this interval.

g) Using your confidence interval, what can you conclude the average life expectancy? Do we have evidence that it is increasing?

h) Your work on #1 depends on a sampling distribution model. Describe the center, shape, and spread of this model.

Answer 1

a)

sample size,n < 5% of population size,N
np>5 and n(1-p)>5

Both are met

b)

Level of Significance,   α =    0.10          
Number of Items of Interest,   x =   229          
Sample Size,   n =    525          
                  
Sample Proportion ,    p̂ = x/n =    0.436          
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0216          
margin of error , E = Z*SE =    1.645   *   0.0216   =   0.0356
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.436   -   0.0356   =   0.4006
Interval Upper Limit = p̂ + E =   0.436   +   0.0356   =   0.4718
                  
90%   confidence interval is (   0.4006   < p <    0.4718   )

c)

We are 90% confident that true mean will fall in above interval

d) Same as above

e)

Ho :   p =    0.46                  
H1 :   p <   0.46       (Left tail test)          
                          
Level of Significance,   α =    0.10                  
Number of Items of Interest,   x =   229                  
Sample Size,   n =    525                  
                          
Sample Proportion ,    p̂ = x/n =    0.4362                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0218                  
Z Test Statistic = ( p̂-p)/SE = (   0.4362   -   0.46   ) /   0.0218   =   -1.0946
                          
critical z value =        -1.282   [excel function =NORMSINV(α)]              
                          
p-Value   =   0.136846443   [excel function =NORMSDIST(z)]              
Decision:   p value>α ,do not reject null hypothesis

                            
There is not enough evidence that   the percentage of females in their labor force is lower than Europe’s rate of 46%

f)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   4.4042
Sample Size ,   n =    20
Sample Mean,    x̅ = ΣX/n =    78.6500

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   19          
't value='   tα/2=   2.0930   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   4.4042   / √   20   =   0.984819
margin of error , E=t*SE =   2.0930   *   0.98482   =   2.061250
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    78.65   -   2.061250   =   76.588750
Interval Upper Limit = x̅ + E =    78.65   -   2.061250   =   80.711250
95%   confidence interval is (   76.59   < µ <   80.71   )

g)

As we can see 77 falls above interval we can say life expectancy has not significantly chnaged

THANKS

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