Question

Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. "Wrinkle recovery angle" measures how well a fabric recovers from wrinkles. Higher is better. Here are data on the wrinkle recovery angle (in degrees) for a random sample of fabric specimens. Assume the populations are approximately normally distributed with unequal variances. A manufacturer believes that the mean wrinkle recovery angle for Hylite is better. A random sample of 11 Hylite (group 1) and 8 Permafresh (group 2) were measured. Test the claim using a 10% level of significance.
Hylite Permafresh
140 124
139 144
138 136
117 133
141 139
102 132
127 141
137 133
164
142
131

Answer 1

Answer:

$t=\frac{134.36-135.25}{\sqrt{\frac{15.71^2}{11}+\frac{6.23^2}{8}}}=-0.17$  

$p_v =P(t_{(17)}$

So the p value is a very high value and using any significance level for example $\alpha=0,1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis and we have enough evidence to support the claim.  Since the mean for Hylite is not significantly less than the mean for Permafresh.

Step-by-step explanation:

1) Data given and notation

Hylite:[140,139, 138 , 117, 141, 102, 127, 137, 164, 142, 131]

Permafresh:[124,144,136,133,139,132,141,133]

$\bar X_{Hylite}$ represent the mean for the sample Hylite

$\bar X_{permafresh}$ represent the mean for the sample LPermafresh

$s_{Hylite}$ represent the sample standard deviation for the sample Stick

$s_{Permafresh}$ represent the sample standard deviation for the sample Liquid

$n_{Hylite}=11$ sample size for the group Hylite

$n_{Permafresh}=8$ sample size for the group Permafresh

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for Hylite is better for Hylie:

H0:$\mu_{Hylite} \geq \mu_{Permafresh}$

H1:$\mu_{Hylite} < \mu_{Permafresh}$

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:

$t=\frac{\bar X_{Hylite}-\bar X_{Perm}}{\sqrt{\frac{s^2_{Hylite}}{n_{Hylite}}+\frac{s^2_{Perm.}}{n_{LPerm.}}}}$ (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.

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)  

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)  

3) Calculate the statistic

First we need to calculate the mean and deviation for each sample, after apply the formulas (2) and (3) we got the following results:

$\bar X_{Hylite}=134.36$ $s_{Hylite}=15.71$

$\bar X_{Permafresh}=135.25$ $s_{Permafresh}=6.23$

And with this we can replace in formula (1) like this:

$t=\frac{134.36-135.25}{\sqrt{\frac{15.71^2}{11}+\frac{6.23^2}{8}}}=-0.17$  

4) Statistical decision

For this case we don't have a significance level provided $\alpha$, but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

$df=n_{Hylite}+n_{Perm}-2=11+8-2=17$

Since is a left tailed test the p value would be:

$p_v =P(t_{(17)}$

So the p value is a very high value and using any significance level for example $\alpha=0,1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis and we have enough evidence to support the claim.  Since the mean for Hylite is not significantly less than the mean for Permafresh.