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.