91Ó°ÊÓ

A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed of the film and believes that this can be achieved by reducing the thickness of the film to 20 mils. Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured. For the 25 -mil film, the sample data result is \(\bar{x}_{1}=1.15\) and \(s_{1}=0.11,\) and for the 20 -mil film the data yield \(\bar{x}_{2}=1.06\) and \(s_{2}=0.09 .\) Note that an increase in film speed would lower the value of the observation in microjoules per square inch. (a) Do the data support the claim that reducing the film thickness increases the mean speed of the film? Use \(\sigma=0.10\), and assume that the two population variances are equal and the underlying population of film speed is normally distributed. What is the \(P\) -value for this test? (b) Find a \(95 \%\) confidence interval on the difference in the two means that can be used to test the claim in part (a).

Step by step solution

01

Define Hypotheses

We're conducting a hypothesis test to determine whether reducing film thickness increases speed. We hypothesize that film speed, as measured by the mean, is higher (or equivalently, lower in microjoules per square inch) for thinner films. Formally, our hypotheses are: \( H_0: \mu_1 = \mu_2 \) and \( H_a: \mu_1 > \mu_2 \), where \( \mu_1 \) is the mean speed for the 25-mil film and \( \mu_2 \) is the mean speed for the 20-mil film.

02

Calculate T-statistic

Given that population variances are assumed to be equal, use the formula for the pooled standard deviation and t-statistic: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \cdot \sqrt{\frac{2}{n}}} \]First, find the pooled standard deviation \( s_p \): \[ s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}} \]Substitute \( n_1 = n_2 = 8 \), \( s_1 = 0.11 \), and \( s_2 = 0.09 \):\[ s_p = \sqrt{\frac{7 \cdot 0.11^2 + 7 \cdot 0.09^2}{8+8-2}} = 0.101 \]Then calculate the t-statistic:\[ t = \frac{1.15 - 1.06}{0.101 \cdot \sqrt{\frac{2}{8}}} = 2.801 \]

03

Determine P-value

With the calculated t-statistic of 2.801 and degrees of freedom (\( df = 8 + 8 -2 = 14 \)), we use a t-distribution table or calculator to find the P-value for a one-tailed test. Using the t-distribution table, the P-value associated with \( t = 2.801 \) and \( df = 14 \) is approximately 0.007, which is less than 0.10, our significance level.

04

Conclusion of Hypothesis Test

Since the P-value (0.007) is less than \( \alpha = 0.10 \), we reject the null hypothesis. This indicates there is sufficient evidence to support the claim that reducing the film thickness to 20 mils increases the mean speed.

05

Calculate Confidence Interval

We calculate a 95% confidence interval for the difference in means using the pooled standard deviation:\[ \bar{x}_1 - \bar{x}_2 \pm t_{\alpha/2} \cdot s_p \sqrt{\frac{2}{n}} \]For a 95% confidence interval, \( t_{\alpha/2} \) for \( df = 14 \) is approximately 2.145. We already have \( \bar{x}_1 - \bar{x}_2 = 0.09 \) and \( s_p = 0.101 \):\[ CI = 0.09 \pm 2.145 \cdot 0.101 \cdot \sqrt{\frac{2}{8}} = 0.09 \pm 0.076 \]Thus, the 95% confidence interval is \([0.014, 0.166]\).

06

Interpret Confidence Interval

The 95% confidence interval for the difference in the means is \([0.014, 0.166]\), which does not include 0. This interval suggests that there is a statistically significant difference in the mean speeds, supporting the claim that reducing thickness increases film speed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval

A confidence interval gives us a range of values within which we expect a population parameter to fall. In the context of this exercise, it helps us estimate the difference between the mean speeds of the two different film thicknesses. To construct a confidence interval, we calculate the range where the true difference in means is likely to fall, with a certain level of confidence (in this case, 95%).

• The calculated difference in sample means is 0.09 microjoules per square inch.
• The critical value from the t-distribution for a 95% confidence level (with 14 degrees of freedom) is approximately 2.145.
• The pooled standard deviation is 0.101, and we use it to compute the standard error for our sample data.
• Our confidence interval calculation becomes \[ CI = 0.09 \pm 2.145 \times 0.101 \times \sqrt{\frac{2}{8}} \]

This provides the interval [0.014, 0.166], meaning we can be 95% confident that the true difference in mean speeds, due to thickness changes, lies in this range. Importantly, this interval does not include zero, indicating a statistically significant difference.

T-statistic

The t-statistic is a value that helps us determine whether the difference in sample means is statistically significant. In this situation, we want to see if the thinner film genuinely results in increased speed (lower energy in microjoules per square inch). When variances are assumed equal, we combine them using a pooled method to compute the t-statistic.

• First, calculate the pooled standard deviation, which accounts for both sample variances and sizes.
• The formula for the t-statistic is \[ t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \cdot \sqrt{\frac{2}{n}}} \]

Here, \[ \bar{x}_1 = 1.15, \ \bar{x}_2 = 1.06, \ s_1 = 0.11, \ s_2 = 0.09, \textrm{and } n = 8.\]Plugging in these numbers, we find that the t-statistic is approximately 2.801, showing that the difference in means is not merely due to random sampling.

Pooled Variance

Pooled variance is an average of the variances from two samples, weighted by their respective sample sizes. It is used when the variances of the two populations are assumed equal. This pooled variance then helps compute the t-statistic.The formula for pooled variance is:\[s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}\]For our case, each film type is sampled eight times, so:

• \( n_1 = n_2 = 8, \ s_1 = 0.11 \), and \( s_2 = 0.09 \)
• With these, the pooled variance becomes: \[ s_p^2 = \frac{7\times 0.11^2 + 7\times 0.09^2}{14} = 0.101^2 \]

The pooled variance tells us how much sample data deviates on average, helping provide an accurate t-statistic for hypothesis testing.

Film Thickness

In this exercise, film thickness plays a crucial role in determining film speed. The engineer hypothesizes that reducing the film thickness from 25 mils to 20 mils could increase film speed. The mean speed of thinner films is observed to be greater, when inversely measured in microjoules per square inch. When there's less material to impede energy movement, the film could indeed perform faster in certain applications.

• The thicker film (25 mils) shows a mean film speed of 1.15.
• The thinner film (20 mils) has a lower mean at 1.06, which corresponds to increased speed.
• Reduction in film thickness could thus lead to better performance as hypothesized.

By quantifying these changes through statistical methods, this experiment supports the hypothesis that a thinner film results in better efficiency.