Show work please In efforts at investigating the cotinine le

Show work please

In efforts at investigating the cotinine levels of smokers, 40 smokers were surveyed who illustrated a mean level of cotinine concentration of 172.5 mg/ml. In this case the population standard deviation s is given as 119.5 mg/ml. ( 44 points )

Employ a 0.01 significance level in order to test the claim that the mean cotinine level of all smokers is 200.0 mg/ml.

Find the associated CI for (2)(a) above.

Find the probability from the P-value for (2)(a) above.

Employ a 0.05 significance level in order to test the claim that the mean cotinine level of all smokers is 152 mg/ml.

Find the associated CI for (2)(d) above.

Find the probability from the P-value for (2)(d) above.

Solution

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   200  
Ha:    u   =/   200  
              
As we can see, this is a    two   tailed test.      
              
Thus, getting the critical z, as alpha =    0.01   ,      
alpha/2 =    0.005          
zcrit =    +/-   2.575829304      
              
Getting the test statistic, as              
              
X = sample mean =    172.5          
uo = hypothesized mean =    200          
n = sample size =    40          
s = standard deviation =    119.5          
              
Thus, z = (X - uo) * sqrt(n) / s =    -1.455441601          
              
Also, the p value is              
              
p =    0.145547047          
              
As |z| < 2.576, and P > 0.01, we   FAIL TO REJECT THE NULL HYPOTHESIS.          

Thus, there is no significant evidence that the mean cotinine level of all smokers is not 200.0 mg/ml. [CONCLUSION]

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b)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    172.5          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    119.5          
n = sample size =    40          
              
Thus,              
Margin of Error E =    48.66928759          
Lower bound =    123.8307124          
Upper bound =    221.1692876          
              
Thus, the confidence interval is              
              
(   123.8307124   ,   221.1692876   ) [ANSWER]

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c)

As we said in part a,

p =    0.145547047   [ANSWER]

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