# Solved (Free): The thickness of a plastic film (in mm) on a substrate material is thought to be influenced by the temperature at which the coating is applied. A completely randomized experiment is carried out

#### ByDr. Raju Chaudhari

Mar 14, 2021

The thickness of a plastic film (in mm) on a substrate material is thought to be influenced by the temperature at which the coating is applied. A completely randomized experiment is carried out. Eleven substrates are coated at 125$^o$C resulting in a sample mean thickness of $\overline{x}_1=103.5$ mm and a sample standard deviation of $s_1=10.2$ mm. Another 13 substrates are coated at 150$^o$C for which $\overline{x}_2 =99.7$ mm and $s_2 = 20.1$ mm are observed. It was originally suspected that raising the process temperature would reduce mean coating thickness.

i) Calculate a 95% confidence interval for the population mean for substrate one.
ii) Calculate a 95% confidence interval for the mean difference in the thickness of the two substrates, is the original claim true (assume the variances are statistically equivalent). Interpret the result.

#### Solution

i) For substrate one, given that sample size $n = 11$, sample mean $\overline{x}_1= 103.5$, sample standard deviation $s_1 = 10.2$.
The confidence level is $1-\alpha = 0.95$.

Specify the formula

$100(1-\alpha)$% confidence interval for the population mean $\mu$ is
 \begin{aligned} \overline{x}_1 - E \leq \mu \leq \overline{x}_1 + E \end{aligned}
where $E = t_{(\alpha/2,n-1)} \frac{s_1}{\sqrt{n}}$, andand $t_{\alpha/2, n-1}$ is the $t$ value providing an area of $\alpha/2$ in the upper tail of the students' $t$ distribution.

Determine the critical value

The critical value of $t$ for given level of significance and $n-1$ degrees of freedom is $t_{\alpha/2,n-1}$.

Thus $t_{\alpha/2,n-1} = t_{0.025,11-1}= 2.228$.

Compute the margin of error

The margin of error for mean is

 \begin{aligned} E & = t_{(\alpha/2,n-1)} \frac{s_1}{\sqrt{n}}\\ & = 2.228 \frac{10.2}{\sqrt{11}} \\ & = 6.852. \end{aligned}

Determine the confidence interval

$95$% confidence interval estimate for population mean is

 \begin{aligned} \overline{x}_1 - E & \leq \mu \leq \overline{x}_1 + E\\ 103.5 - 6.852 & \leq \mu \leq 103.5 + 6.852\\ 96.648 &\leq \mu \leq 110.352. \end{aligned}

Thus, $95$% confidence interval estimate for population mean for substrate one is $(96.648,110.352)$.

ii) Given that $n_1 = 11$, $\overline{x}_1 =103.2$, $s_1 = 10.2$, $n_2 =13$, $\overline{x}_2 =99.7$ and $s_2 = 20.1$.

We want to determine $95$% confidence interval estimate for the difference $(\mu_1-\mu_2)$.

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

Specify the formula

$100(1-\alpha)$% confidence interval estimate for the difference $(\mu_1-\mu_2)$ is

 \begin{aligned} (\overline{x}_1 -\overline{x}_2)- E & \leq (\mu_1-\mu_2) \leq (\overline{x}_1 -\overline{x}_2) + E. \end{aligned}
where $E = t_{\alpha/2,n_1+n_2-2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$ and $t_{\alpha/2, n_1+n_2-2}$ is the $t$ value providing an area of $\alpha/2$ in the upper tail of the students' $t$ distribution.

Determine the critical value

The critical value $t_{\alpha/2,n_1+n_2-2} = t_{0.025,22} = 2.074$.

Compute the margin of error

The margin of error for difference of means $\mu_1-\mu_2$ is

 \begin{aligned} E & = t_{\alpha/2,n_1+n_2-2} \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ & = 2.074 \sqrt{\frac{10.2^2}{11}+\frac{20.1^2}{13}}\\ & = 13.205. \end{aligned}

Determine the confidence interval

$95$% confidence interval estimate for the difference $(\mu_1-\mu_2)$ is

 \begin{aligned} (\overline{x}_1 -\overline{x}_2)- E & \leq (\mu_1-\mu_2) \leq (\overline{x}_1 -\overline{x}_2) + E\\ (103.2-99.7) - 13.205 & \leq (\mu_1-\mu_2) \leq (103.2-99.7) + 13.205\\ -9.705 & \leq (\mu_1-\mu_2) \leq 16.705. \end{aligned}

Thus, $95$% confidence interval estimate for the difference $(\mu_1-\mu_2)$ is $(-9.705,16.705)$.

As the interval include the value 0, we conclude that the claim of that raising the process temperature would reduce mean coating thickness is not acceptable.