The concentration of active ingredient in a liquid laundry detergent is thought to be affected by the type of catalyst used in the process. The standard deviation of active concentration is known to be 3 grams per liter, regardless of the catalyst type. Ten observations on concentration are taken with each catalyst, and the data follow:

Catalyst 1: 57.5, 66.4, 65.4, 65.4, 65.2, 62.6, 67.6, 63.7, 67.2, 71.0

Catalyst 2: 67.0, 72.4, 70.3, 69.3, 64.8, 69.6, 68.6, 69.4, 65.3, 68.8

(a) Find a 95% two-sided confidence interval on the difference in mean active concentrations for the two catalysts. Find the P-value.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X_1 =65.2[/tex] represent the sample mean 1

[tex]\bar X_2 =68.55[/tex] represent the sample mean 2

n1=10 represent the sample 1 size

n2=10 represent the sample 2 size

[tex]s_1 =3.55[/tex] sample standard deviation for sample 1

[tex]s_2 =2.29[/tex] sample standard deviation for sample 2

[tex]\sigma =3[/tex] represent the population standard deviation

[tex]\mu_1 -\mu_2[/tex] parameter of interest.

Part a

The confidence interval for the difference of means is given by the following formula:

[tex](\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\sigma^2(\frac{1}{n_1}+\frac{1}{n_2})}[/tex] (1)

The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:

[tex]\bar X_1 -\bar X_2 =65.2-68.55=-3.35[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]

The standard error is given by the following formula:

[tex]SE=\sqrt{\sigma^2(\frac{1}{n_1}+\frac{1}{n_2})}[/tex]

And replacing we have:

[tex]SE=\sqrt{3^2(\frac{1}{10}+\frac{1}{10})}=1.342[/tex]

Confidence interval

Now we have everything in order to replace into formula (1):

[tex]-3.35-1.96\sqrt{3^2(\frac{1}{10}+\frac{1}{10})}=-5.980[/tex]

[tex]-3.35+1.96\sqrt{3^2(\frac{1}{10}+\frac{1}{10})}=-0.720[/tex]

So on this case the 95% confidence interval would be given by [tex]-5.980 \leq \mu_1 -\mu_2 \leq -0.720[/tex]

If the system of hypothesis are given by:

Null Hypothesis:[tex]\mu_1 -\mu_2=0[/tex]

Alternative hypothesis:[tex]\mu_1 -\mu_2 \neq 0[/tex]

The statistic would be:

[tex]z=\frac{\bar X_1 -\bar X_2 -0}{\sqrt{\sigma^2(\frac{1}{n_1}+\frac{1}{n_2})}}[/tex]

And if we replace we got:

[tex]z=\frac{65.2 -68.55 }{\sqrt{3^2(\frac{1}{10}+\frac{1}{10})}}=-2.497[/tex]

And the p value would be given by:

[tex]p_v =2*P(Z<-2.497)=0.012[/tex]

And with 5% of significance we have enough evidence to reject the null hypothesis since the [tex]p_v < \alpha[/tex]