Question

To study the effect of curing temperature on shear strength of a certain rubber compound, 70 specimens were cured at 150°C and 90 were cured at 130°C. The specimens cured at 150°C had an average shear strength of 620 psi, with a standard deviation of 20 psi. Those cured at 130°C had an average shear strength of 750 psi, with a standard deviation of 30 psi. Let μX represent the population mean strength for the specimens cured at 130°C and let μY represent the population mean strength for the specimens cured at 150°C. Find a 95% confidence interval for the difference μX−μY. Round the answers to two decimal places.

Answer 1

Answer:

[122.23, 137.77]

Step-by-step explanation:

In order to do this, we will be using the two sample z-statistic  given by the formula

$\large z=\frac{(\bar x_1 -\bar x_2)-(\mu_1-\mu_2)}{\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}}$

where

$\large \bar x_1$ = the sample mean strength for the specimens cured at 130°C

$\large \bar x_2$ = the sample mean strength for the specimens cured at 150°C

$\large \mu_1$ =  the population mean strength for the specimens cured at 130°C (μX)

$\large \mu_2$ = the population mean strength for the specimens cured at 150°C (μY)

$\large \sigma_1$ = the sample standard deviation for the specimens cured at 130°C

$\large \sigma_2$ = the sample standard deviation for the specimens cured at 150°C

$\large n_1$ = the sample size for the specimens cured at 130°C

$\large n_2$ = the sample size for the specimens cured at 150°C

z = 1.96, the z-score for a 95% confidence interval

Replacing values we have

$\large 1.96=\frac{(750-620)-(\mu_1-\mu_2)}{\sqrt{30^2/90+20^2/70}}$

$\large 1.96=\frac{130-(\mu_1-\mu_2)}{\sqrt{10+5.714285}}=\frac{130-(\mu_1-\mu_2)}{3.9641}$

So, the 95% confidence interval for μX- μY = $\large \mu_1-\mu_2$ is the interval

[130 - 1.96*3.9641, 130 +1.96*3.9641] = [122.23, 137.77] rounded to two decimal places.