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 89 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 (b) find a 99% confidence interval on the difference in mean yields that can be used to test the claim in part (a). round your answer to two decimal places (e. g. 98.76).

a:  Yes

b:  -6.688 < µ1- µ2 < -1.312  

Step-by-step explanation:

We need to conduct a hypothesis test for the difference of 2 means, then construct a confidence interval for the difference between those 2 mean.

The hypothesis test:

The null and alternate hypothesis for this test are

H0:  µ1 - µ2 = 0    

H1:  µ1 - µ2 < 0  (claim)

We are testing whether catalyst 2 produces a higher yield.  We always test for equality, that's why the null has an equals sign.  If catalyst 2 has a higher yield, the difference would result in a negative a number (take a number, and subtract a larger number than that number).

The problem just asks if there is evidence to support the claim that catalyst 2 produces a higher yield, it doesn't give a significance level, so we will use the same significance level for part b, which is 99%.  We will be using a t-score since both samples sizes are less than 30.  For this, the degress of freedom are 2 less than the sum of the two sample sizes. in this case 25.  This is a left tailed test since we are testing for a difference, claiming that µ2 is greater, so if the test statistic is too low a value, we will reject the null.  

The corresponding t-score for a significance level of 99% with a one tailed test is t < -2.485

See attached photo 1 for the hypothesis test calculations...  

Our test statistic is t = -4.148

-4.148 < -2.485, so we reject the null hypothesis.  This means that there is sufficient evidence to support the claim that catalyst 2 produces a higher yield that catalyst 1.

We must now build a confidence interval.  See attached photo 2 for the construction of the confidence interval

f(x) = y

f(-3) = -5

step-by-step explanation:

step-by-step explanation:

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