Question

Consider the accompanying data on breaking load (kg/25mm width) for various fabrics in both an unabraded condition and an abraded condition.
Fabric

1 2 3 4 5 6 7 8

U 36.4 55.0 51.5 38.7 43.2 48.8 25.6 49.8

A 28.5 20.0 46.0 34.5 36.5 52.5 26.5 46.5


a. Test if the average breaking load of various fabrics in unabraded condition is greater than that in abraded condition using the rejection method at significance level 0.01.

b. Compute a 99% confidence interval for the true average difference of breaking load for fabric in unabraded and abraded condition.

Answer 1

Answer:

a) $t=\frac{(43.625-36.375)-0}{\sqrt{\frac{9.694^2}{8}+\frac{11.253^2}{8}}}}=1.38$

 $p_v = P(t_{14}>1.38) =0.0946$

Since the p value for this case is higher than the significance level of 0.01 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that we don't have significant differences between the true means

b) $(43.625 -36.375) - 2.98 \sqrt{\frac{9.694^2}{8} +\frac{11.253^2}{8}} = -23.976$

$(43.625 -36.375) + 2.98 \sqrt{\frac{9.694^2}{8} +\frac{11.253^2}{8}} = 38.476$

Step-by-step explanation:

We have the following data given:

U 36.4 55.0 51.5 38.7 43.2 48.8 25.6 49.8

A 28.5 20.0 46.0 34.5 36.5 52.5 26.5 46.5

$\bar X_{U}=43.625$ represent the mean for sample U

$\bar X_{A}=36.375$ represent the mean for sample A

$s_{U}=9.694$ represent the sample standard deviation for U

$s_{A}=11.253$ represent the sample standard deviation for A

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

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

$\alpha=0.01$ Significance level provided

t would represent the statistic

Part a

For this case we want to try test if the average breaking load of various fabrics in unabraded condition is greater than that in abraded, so then the hypothesis are:

Null hypothesis:$\mu_{U}-\mu_{A} \leq 0$  

Alternative hypothesis:$\mu_{U} - \mu_{A}> 0$  

The statistic for this case is given by:

$t=\frac{(\bar X_{U}-\bar X_{A})-\Delta}{\sqrt{\frac{s^2_{U}}{n_{U}}+\frac{s^2_{A}}{n_{A}}}}$ (1)

And the degrees of freedom are given by :

$df = n_U + n_A -2= 8+8-2 =14$

The statistic for this case would be:

$t=\frac{(43.625-36.375)-0}{\sqrt{\frac{9.694^2}{8}+\frac{11.253^2}{8}}}}=1.38$

P value

The p value can be calculated like this:

$p_v = P(t_{14}>1.38) =0.0946$

Since the p value for this case is higher than the significance level of 0.01 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that we don't have significant differences between the true means

Part b

The confidence interval for the true average difference would be given by:

$(\bar X_U -\bar X_A) \pm t_{\alpha/2} \sqrt{\frac{s^2_U}{n_U} +\frac{s^2_A}{n_A}}$

And if we find the critical value for this case we got:

$t_{\alpha/2}= 2.98$

And replacing we got:

$(43.625 -36.375) - 2.98 \sqrt{\frac{9.694^2}{8} +\frac{11.253^2}{8}} = -23.976$

$(43.625 -36.375) + 2.98 \sqrt{\frac{9.694^2}{8} +\frac{11.253^2}{8}} = 38.476$