Drug content assessment. Refer to Exercise 8.16 (p. 467)and the Analytical Chemistry (Dec. 15, 2009) study in which scientists used high-performance liquid chromatography to determine the amount of drug in a tablet. Recall that 25 tablets were produced at each of two different, independent sites. The researchers want to determine if the two sites produced drug concentrations with different variances. A Minitab printout of the analysis follows. Locate the test statistic and p-value on the printout. Use these values $\alpha =.05$and to conduct the appropriate test for the researchers.

Test and CI for two Variances: Content vs Site

Method

Null hypothesis $\raisebox{1ex}{$\alpha \left(1\right)$}\!\left/ \!\raisebox{-1ex}{$\alpha \left(2\right)=1$}\right.$

Alternative hypothesis $\raisebox{1ex}{$\alpha \left(1\right)$}\!\left/ \!\raisebox{-1ex}{$\alpha \left(2\right)\ne $}\right.1$

F method was used. This method is accurate for normal data only.

Statistics

Site N St Dev Variance 95% CI for St Devs

1 25 3.067 9.406 (2.195,4.267)

2 25 3.339 11.147 (2.607,4.645)

Ratio of standard deviation =0.191

Ratio of variances=0.844

95% Confidence Intervals

Method CI for St Dev Ratio CI Variance Ratio

F (0.610, 1.384) (0.372, 1.915)

Tests

Method DF1 DF2 Test statistic p-value

F 24 24 0.84 0.681

$$\begin{gathered} n=25 \\ \alpha =.05 \end{gathered}$$

Finding the test statistic and p-value on the printout is required for this question. Additionally, you must conduct the test using the alpha value.

Individuals must choose the proper test because conclusions, interpretations, and suggestions will be based on our analyses.

1.T-tests

Comparing two variables within the same population is done using paired t-tests. For instance, obtained the same group's pre- and post-scores or scores from the same group under various circumstances.

The test aims to determine whether statistical support exists for the claim that the mean difference between observations for a given outcome differs significantly from zero. These t-tests are also referred to as dependent t-tests or repeated measures t-tests.

The variable of interest has to be continuous, the respondents in each situation should be the same, must usually distribute distinction between the pair measures, and your data must not contain any outliers to use this test.

2.Anova-test

The t-tests mentioned above have a few drawbacks. Only two means can be compared, and they can only be applied to one independent variable. ANOVA, which compares various means, is a statistical method that can be used when more than one independent variable has been changed.

3.F-test

An F-test is any statistical test that produces a test statistic with an F-distribution under the null hypothesis. It is frequently used when contrasting models fitted to data sets to determine which statistical model best represents the population from which sampled the data. After the models have been least squares fitted to the data, exact "F-tests" are typically needed.

Test and CI for two Variances: Content vs Site

Method

Null hypothesis $\raisebox{1ex}{$\alpha \left(1\right)$}\!\left/ \!\raisebox{-1ex}{$\alpha \left(2\right)=1$}\right.$

Alternative hypothesis $\raisebox{1ex}{$\alpha \left(1\right)$}\!\left/ \!\raisebox{-1ex}{$\alpha \left(2\right)\ne 1$}\right.$

F method was used. This method is accurate for normal data only.

Statistics

Site N St Dev Variance 95% CI for St Devs

1 25 3.067 9.406 (2.195,4.267)

2 25 3.339 11.147 (2.607,4.645)

Ratio of standard deviation =0.191

Ratio of variances=0.844

95% Confidence Intervals

Method CI for St Dev Ratio CI Variance Ratio

F (0.610, 1.384) (0.372, 1.915)

Tests

Method DF1 DF2 Test statistic p-value

F 24 24 0.84 0.681

The printout displays this F-test. The printout highlights the test statistic (F = 0.84) and p-value (p-value = 0.681). No need to reject the null hypothesis that the population variances of the success indices are not equal because a = 0.05 is less than the p-value.

Let

Null hypothesis $\raisebox{1ex}{$\alpha \left(1\right)$}\!\left/ \!\raisebox{-1ex}{$\alpha \left(2\right)$}\right.=1$

Alternative hypothesis $\raisebox{1ex}{$\alpha \left(1\right)$}\!\left/ \!\raisebox{-1ex}{$\alpha \left(2\right)$}\right.\ne 1$

Here need to ensure that the upper tail is used when forming the rejection region for the two-tailed F-test and include the more considerable sample variance in the F-test statistic's numerator to achieve this.

For example, the numerator localid="1664862541925" $S^{2}1$and denominator ${S^2}_2$have $$\begin{gathered} df=n-1 \\ df=25-1 \\ df=24 \end{gathered}$$, respectively. The test statistic will be a result

$$\begin{gathered} F=\frac{L\arg er sample variance}{Smaller sample variance} \\ F=\frac{S^{2}1}{S^{2}2} \\ F=\frac{3.067}{3.339} \\ F=0.918 \end{gathered}$$

When the calculated value of F exceeds the tabulated value, reject$\raisebox{1ex}{$\alpha \left(1\right)$}\!\left/ \!\raisebox{-1ex}{$\alpha \left(2\right)$}\right.=1$

for$\alpha =0.05$

$$\begin{gathered} F_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=0.025 \\ {F}_{\alpha }=1.19 \end{gathered}$$

F = 0.918 is in the region of rejection. Therefore the test indicates that the information is sufficient to show that the population variances are different.