Question

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.

Answer 1

Answer:

The 95% confidence interval would be given by $-5.980 \leq \mu_1 -\mu_2 \leq -0.720$  

$p_v =2*P(Z$

Step-by-step explanation:

Previous concepts  

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".  

$\bar X_1 =65.2$ represent the sample mean 1

$\bar X_2 =68.55$ represent the sample mean 2

n1=10 represent the sample 1 size  

n2=10 represent the sample 2 size  

$s_1 =3.55$ sample standard deviation for sample 1

$s_2 =2.29$ sample standard deviation for sample 2

$\sigma =3$ represent the population standard deviation

$\mu_1 -\mu_2$ parameter of interest.

Part a

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

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

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =65.2-68.55=-3.35$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, 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 $z_{\alpha/2}=1.96$  

The standard error is given by the following formula:

$SE=\sqrt{\sigma^2(\frac{1}{n_1}+\frac{1}{n_2})}$

And replacing we have:

$SE=\sqrt{3^2(\frac{1}{10}+\frac{1}{10})}=1.342$

Confidence interval

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

$-3.35-1.96\sqrt{3^2(\frac{1}{10}+\frac{1}{10})}=-5.980$  

$-3.35+1.96\sqrt{3^2(\frac{1}{10}+\frac{1}{10})}=-0.720$  

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

If the system of hypothesis are given by:

Null Hypothesis:$\mu_1 -\mu_2=0$

Alternative hypothesis:$\mu_1 -\mu_2 \neq 0$

The statistic would be:

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

And if we replace we got:

$z=\frac{65.2 -68.55 }{\sqrt{3^2(\frac{1}{10}+\frac{1}{10})}}=-2.497$

And the p value would be given by:

$p_v =2*P(Z$

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