Question

Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yield of 85 and a sample standard deviation of 3. Fifteen batches were prepared using catalyst 2, and they resulted in an average yield of 91 with a standard deviation of 2. Assume that yield measurements are approximately normally distributed with the same standard deviation.(a) Is there evidence to support the claim that catalyst 2 produces higher mean yield than catalyst 1? Use alpha = 0.01.(b) Find a 99% confidence interval on the difference in mean yields that can be used to test the claim in part (a). (e.g. 98.76). mu_1 - mu_2 lessthanorequalto the tolerance is +/-2%

Answer 1

Answer:

a) $t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$  

$df=12+15-2=25$  

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) $(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309$

$(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691$

Step-by-step explanation:

Notation and hypothesis

When we have two independent samples from two normal distributions with equal variances we are assuming that  

$\sigma^2_1 =\sigma^2_2 =\sigma^2$  

And the statistic is given by this formula:  

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}$  

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:  

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$  

This last one is an unbiased estimator of the common variance $\sigma^2$  

Part a

The system of hypothesis on this case are:  

Null hypothesis: $\mu_2 \leq \mu_1$  

Alternative hypothesis: $\mu_2 > \mu_1$  

Or equivalently:  

Null hypothesis: $\mu_2 - \mu_1 \leq 0$  

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

Our notation on this case :  

$n_1 =12$ represent the sample size for group 1  

$n_2 =15$ represent the sample size for group 2  

$\bar X_1 =85$ represent the sample mean for the group 1  

$\bar X_2 =91$ represent the sample mean for the group 2  

$s_1=3$ represent the sample standard deviation for group 1  

$s_2=2$ represent the sample standard deviation for group 2  

First we can begin finding the pooled variance:  

$\S^2_p =\frac{(12-1)(3)^2 +(15 -1)(2)^2}{12 +15 -2}=6.2$  

And the deviation would be just the square root of the variance:  

$S_p=2.490$  

Calculate the statistic

And now we can calculate the statistic:  

$t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$  

Now we can calculate the degrees of freedom given by:  

$df=12+15-2=25$  

Calculate the p value

And now we can calculate the p value using the altenative hypothesis:  

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

Conclusion

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis.

Part b

For this case the confidence interval is given by:

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

For the 99% of confidence we have $\alpha=1-0.99 = 0.01$ and $\alpha/2 =0.005$ and the critical value with 25 degrees of freedom on the t distribution is $t_{\alpha/2}= 2.79$

And replacing we got:

$(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309$

$(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691$