Question

A polymer is manufactured in a batch chemical process. Viscosity measurements are normally made on each batch, and long experience with the process has indicated that the variability in the process is fairly stable with o 20. Fifteen batch

viscosity measurements are given as follows:
724, 718, 776, 760, 745, 759, 795, 756, 742, 740, 761, 749, 739, 747, 742
A process change that involves switching the type of catalyst used in the process is made. Following the process change, eight batch viscosity measurements are taken:
735, 775, 729, 755, 783, 760, 738, 780
Assume that process variability is unaffected by the catalyst change. If the difference in mean batch viscosity is 10 or less, the manufacturer would like to detect it with a high probability.
(a) Formulate and test an appropriate hypothesis using a 0.10. What are your conclusions? Find the P-value.
(b) Find a 9096 confidence interval on the difference in mean batch viscosity resulting from the process change.
(c) Compare the results of parts (a) and (b) and discuss your findings.

Answer 1

a) Null hypothesis:
`\mu_(s)-\mu_(n)\geq 10`

Alternative hypothesis:
`\mu_(s) - \mu_(n)<10`

`p_v =P(Z<-1.904)=0.0284`

Comparing the p value with the significance level
`\alpha=0.1` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly lower than 10.

b) The 90% confidence interval would be given by
`-21.035 \leq \mu_1 -\mu_2 \leq 7.685`

c) If we analyze the interval obtained we see that the interval contains the value of -10 so then we agree with the result obtained from the hypothesis, that he alternative hypothesis is true.

Explanation:

Data given and notation

`\bar X_(s)=750.2` represent the mean for the sample standard process

`\bar X_(n)=756.875` represent the mean for the sample with the new process

`s_(s)=19.128` represent the sample standard deviation for the sample standard process

`s_(n)=21.283` represent the sample standard deviation for the new process

`\sigma_s=\sigma_n=\sigma=20` represent the population standard deviation for both samples.

`n_(s)=15` sample size for the group Cincinnati

`n_(n)=8` sample size for the group Pittsburgh

z would represent the statistic (variable of interest)

Concepts and formulas to use

(a) Formulate and test an appropriate hypothesis using a 0.10. What are your conclusions?

We need to conduct a hypothesis in order to check if the difference in mean batch viscosity is 10 or less system of hypothesis would be:

Null hypothesis:
`\mu_(s)-\mu_(n)\geq 10`

Alternative hypothesis:
`\mu_(s) - \mu_(n)<10`

We have the population standard deviation, so for this case is better apply a z test to compare means, and the statistic is given by:

`z=\frac{(\bar X_(s)-\bar X_(n))-\Delta}{\sqrt{(\sigma^2_(s))/(n_(s))+(\sigma^2_(n))/(n_(n))}}` (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

With the info given we can replace in formula (1) like this:

`z=\frac{(750.2-756.875)-10}{\sqrt{(20^2)/(15)+(20^2)/(8)}}}=-1.904`

Statistical decision

Since is a one tail left test the p value would be:

`p_v =P(Z<-1.904)=0.0284`

Comparing the p value with the significance level
`\alpha=0.1` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly lower than 10.

(b) Find a 90% confidence interval on the difference in mean batch viscosity resulting from the process change.

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

`(\bar X_s -\bar X_n) \pm z_(\alpha/2)\sqrt{\sigma^2((1)/(n_s)+(1)/(n_s))}` (1)

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

`\bar X_s -\bar X_n =750.2-756.875=-6.675`

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that
`z_(\alpha/2)=1.64`

The standard error is given by the following formula:

`SE=\sqrt{\sigma^2((1)/(n_s)+(1)/(n_n))}`

And replacing we have:

`SE=\sqrt{20^2((1)/(15)+(1)/(8))}=8.756`

Confidence interval

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

`-6.675-1.64\sqrt{20^2((1)/(15)+(1)/(8))}=-21.035`

`-6.675+1.64\sqrt{20^2((1)/(15)+(1)/(8))}=7.685`

So on this case the 90% confidence interval would be given by
`-21.035 \leq \mu_1 -\mu_2 \leq 7.685`

(c) Compare the results of parts (a) and (b) and discuss your findings.

If we analyze the interval obtained we see that the interval contains the value of -10 so then we agree with the result obtained from the hypothesis, that he alternative hypothesis is true.