Question

A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed of the film, and believes that this can be achieved by reducing the thickness of the film to 20 mils. Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured. For the 25-mil film the sample data result is x1 = 1.17 and s1 = 0.11 , while for the 20-mil film, the data yield x2 = 1.04 and s2 = 0.09 . Note that an increase in film speed would lower the value of the observation in microjoules per square inch.

Do the data support the claim that reducing the film thickness increases the mean speed of the film? Use alpha=0.05

Answer 1

`t=\frac{1.17-1.04}{\sqrt{(0.11^2)/(8)+(0.09^2)/(8)}}}=2.587`

`df=n_(1)+n_(2)-2=8+8-2=14`

Since is a one sided test the p value would be:

`p_v =P(t_((14))>2.587)=0.0108`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, we have enough evidence to reject the null hypothesis on this case and the 25 mil film have a mean greater than the 20 mil film so then the claim is not appropiate

Explanation:

Data given and notation

`\bar X_(1)=1.17` represent the mean for the sample 1 (25 mil film)

`\bar X_(2)=1.04` represent the mean for the sample 2 (20 mil film)

`s_(1)=0.11` represent the sample standard deviation for the sample 1

`s_(2)=0.09` represent the sample standard deviation for the sample 2

`n_(1)=8` sample size selected for 1

`n_(2)=8` sample size selected for 2

`\alpha=0.05` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if reducing the film thickness increases the mean speed of the film, the system of hypothesis would be:

Null hypothesis:
`\mu_(1) \leq \mu_(2)`

Alternative hypothesis:
`\mu_(1) > \mu_(2)`

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

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

t-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

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

`t=\frac{1.17-1.04}{\sqrt{(0.11^2)/(8)+(0.09^2)/(8)}}}=2.587`

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n_(1)+n_(2)-2=8+8-2=14`

Since is a one sided test the p value would be:

`p_v =P(t_((14))>2.587)=0.0108`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, we have enough evidence to reject the null hypothesis on this case and the 25 mil film have a mean greater than the 20 mil film so then the claim is not appropiate