Question

A random sample of 25 Tide powder showed a mean content of 237 mg with a variance
of 8.56 mg , while a sample of 20 Ariel powder detergent showed a mean content of 240
mg with a variance of 9.75 mg. Using a .05 level, is there a difference in the mean
content of the two brands of powder detergent?
Use: Step-wise method (Take note that clf is (25+20)-2=43.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1:\mu_1\not=\mu_2$

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are  $F_L=0.426$ and $F_U=2.452,$ and since $F=0.7708,$ then the null hypothesis of equal variances is not rejected.

We assume that the population variances are equal, so then the number of degrees of freedom is computed as follows:

$$df=n_1+n_2-2=25+20-2=43$$

It is found that the critical value for this two-tailed test is $t_c=2.017,$ for $\alpha=0.05$

and $df=43.$  

The rejection region for this two-tailed test is $R=\{t:|t|>2.017\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{X_1}-\bar{X_2}}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$ $$=-0.716$$

$$=\dfrac{237-240}{\sqrt{\dfrac{(25-1)(8.56)^2+(20-1)(9.75)^2}{25+20-2}(\dfrac{1}{25}+\dfrac{1}{20})}}$$

$$=-1.098$$

Since it is observed that $|t|=1.098<2.017=t_c,$  it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that population mean $\mu_1$ is greater than $\mu_2,$ at the 0.05 significance level.

Therefore, there is not enough evidence to claim that there is a difference in the mean

content of the two brands of powder detergent, at the 0.05 significance level.

Using the P-value approach:

The sign of a t-value tells us the direction of the effect , which has no bearing on the significance of the difference between groups. Therefore, it is common to report the t-value as the absolute value of the t-value given by the statistics program.

From t-tables for $df=43$ and $t=1.098$ we have p-value $p=2\cdot P(t<-1.098)=0.2783,$ and since $p=0.2783>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that population mean $\mu_1$ is greater than $\mu_2,$ at the 0.05 significance level.

Therefore, there is not enough evidence to claim that there is a difference in the mean

content of the two brands of powder detergent, at the 0.05 significance level.