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

a)
`t=\frac{(91-85)-(0)}{2.490\sqrt{(1)/(12)}+(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{(1)/(12) +(1)/(15)} =3.309`

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

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{(1)/(n_1)}+(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 =((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 =((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{(1)/(12)}+(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{(1)/(n_1) +(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{(1)/(12) +(1)/(15)} =3.309`

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