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.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators