Question

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

Answer 1

Given :

`$\mu = 24$`

n = 15

`$\overline x = 24.89$`

s = 0.5902

The hypothesis :

`$H_0 : \mu = 25$`

`$H_a : \mu \\eq 25$`

This is a 2 tailed test.

The significance level is
`$95 \% \ ( \alpha =0.05)$`

The test statistic :

`$Z=(\overline x - \mu)/((\sigma)/(\sqrt n))$`

`$Z=(24.89-25)/((0.5902)/(\sqrt 15))$`

= -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
`$Z_(observed) > Z_(critical)$` or if
`$Z_(observed) < -Z_(critical)$`, then reject
`$H_0$`.

Also if p value is less than α, then reject
`$H_0$`.

The decision :

Since the Z falls in between +1.96 and -1.96, we fail to reject the
`$H_0$`. Also since p value is greater than α, we fail to reject
`$H_0$`.

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 :

`$\text{Mean} = \frac{\text{sum of observation}}{\text{total observations}}$`

Standard deviation =
`$\sqrt{\text{variance}}$`

Variance =
`$\frac{\text{sum of squares(SS)}}{n-1}$`

Where, SS =
`$\sum (X - \text{mean})^2$`

X Mean
`$(X-\text{mean})^2$`

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 =
`$(SS)/(n-1)$` 0.348392857

Standard deviation 0.5902