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xenn [34]
3 years ago
13

Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yi

eld 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%
Mathematics
1 answer:
aalyn [17]3 years ago
7 0

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 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 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

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