The Westbrook Pharmaceutical Company manufactures atorvastatin pills, designed to lower cholesterol levels. Listed below are the amounts (in mg) of atorvastatin in a random sample of the pills. Use a 0.05 significance level to test the claim that the pills come from a population in which the amount of atorvastatin is equal to 25mg. 24.1, 24.4, 24.3, 24.9, 24.1, 26.2, 25.1, 24.7, 24.4, 25.0, 24.7, 25.1, 25.3, 25.5, 25.5

Solution :

Given :

n = 15

s = 0.5902

The hypothesis :

This is a 2 tailed test.

The significance level is

The test statistic :

   = -0.72

The p value : The p value for Z = -0.72 is 0.4716

The critical value : the critical value at α = 0.05 is +1.96 to -1.96

The decision rule :

If or if , then reject .

Also if p value is less than α, then reject .

The decision :

Since the Z falls in between +1.96 and -1.96, we fail to reject the . Also since p value is greater than α, we fail to reject  .

The conclusion :

There is not sufficient evidence at the 95% significance level to warrant rejection of the claim that the pills come from a population in which the amount of the atorvastatin is equal to 25 mg.

Now calculating the mean and the standard deviation :

Standard deviation =

Variance =

Where, SS =

X                   Mean                

24.1               24.89                    0.62

24.4              24.89                     0.24

24.3              24.89                    0.35

24.9             24.89                       0

24.1              24.89                    0.62

24.2             24.89                    1.72

24.1              24.89                    0.04

26.2             24.89                    0.04

25.1              24.89                   0.24

25                24.89                    0.01

24.7             24.89                    0.04

25.1             24.89                    0.04

25.3             24.89                   0.17

25.5             24.89                   0.37

25.5            24.89                   0.37

n                                                   15

Sum                                           373.3

Average                                    24.89

SS                                              4.8775

Variance =                       0.348392857

Standard deviation                    0.5902      

she was supposed to add the hours(38.5) and multiply the amount of money she earned(7.50) resulting in $288.75 .

correct me if i'm wrong and if it is i'm truly sorry.