/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 A photoconductor film is manufac... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(20-\mathrm{mil}\) 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).

Short Answer

Expert verified
Results support the claim; P-value ≈ 0.01. Confidence interval: [0.014, 0.166].

Step by step solution

01

Identify Hypotheses

First, we need to define the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). The claim that reducing the film thickness increases the mean speed implies a comparison of means. Since speed is inversely related to the measured value, we want to verify if the thinner film has a lower average value. Thus, our hypotheses are:\[ H_0: \mu_1 = \mu_2 \] \[ H_a: \mu_1 > \mu_2 \]where \(\mu_1\) and \(\mu_2\) are the mean speeds for the 25-mil and 20-mil films, respectively.
02

Calculate the Test Statistic

We will use a two-sample t-test to compare the means, assuming equal variances. The test statistic \( t \) can be calculated as follows:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{2}{n}}} \]where \[ s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}} \] is the pooled standard deviation. Given that \( n_1 = n_2 = 8 \), \( \bar{x}_1 = 1.15 \), \( \bar{x}_2 = 1.06 \), \( s_1 = 0.11 \), and \( s_2 = 0.09 \), compute \( s_p = \sqrt{\frac{7 \cdot 0.11^2 + 7 \cdot 0.09^2}{14}} = 0.10 \). Thus, the test statistic is:\[ t = \frac{1.15 - 1.06}{0.10 \cdot \sqrt{\frac{2}{8}}} = 2.54 \].
03

Find the P-Value

To find the P-value, compare the test statistic \( t = 2.54 \) with the t-distribution with \( n_1 + n_2 - 2 = 14 \) degrees of freedom. This is a one-tailed test because we are testing for an increase in speed (decrease in mean value). A t-value of 2.54 with 14 degrees of freedom will give a P-value less than \(0.05\) (specifically around \(0.01\)), indicating that we reject \( H_0 \) with a significance level \(\alpha = 0.05\).
04

Calculate Confidence Interval

To find the 95% confidence interval for the difference in means, use the formula:\[ (\bar{x}_1 - \bar{x}_2) \pm t* \cdot s_p \cdot \sqrt{\frac{2}{n}} \]With \(t*=2.145\) for a 95% two-tailed test and 14 degrees of freedom, compute:\[ (1.15 - 1.06) \pm 2.145 \cdot 0.10 \cdot \sqrt{\frac{2}{8}} \]\[ 0.09 \pm 0.076 \]Thus, the confidence interval is \([0.014, 0.166]\). Because the interval does not include 0, this supports the claim that the means are significantly different.

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Key Concepts

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

t-test
In statistical hypothesis testing, the **t-test** is used to determine if there is a significant difference between the means of two groups. It is especially useful when dealing with small sample sizes and when the population standard deviation is unknown. The t-test assumes that the data follows a normal distribution and that the variances of the two groups are equal.
A t-test can take several forms, including:
  • **Independent two-sample t-test:** Compares the means of two independent groups.
  • **Paired sample t-test:** Compares means from the same group at different times.
  • **One-sample t-test:** Determines if the sample mean differs from a known or hypothesized population mean.
In our exercise, we use an independent two-sample t-test to compare film speeds for two different thicknesses. We calculate the t-statistic, which measures how many standard deviations the observed difference in sample means is from the hypothesized population difference. The formula is:\[t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{2}{n}}}\]where \(s_p\) is the pooled standard deviation. A calculated t-value helps determine the p-value, which tells us whether to reject the null hypothesis.
If the calculated p-value is less than the chosen significance level (often 0.05), we reject the null hypothesis, suggesting a significant difference exists between the group means.
confidence interval
A **confidence interval** provides a range of values that is likely to contain the true difference between two population means. It is a way to express the uncertainty or potential error in a statistical estimate. The confidence level, often expressed as a percentage, indicates how sure we are that the interval contains the true difference in the population means.
To calculate the confidence interval for two independent samples, we use the formula:\[(\bar{x}_1 - \bar{x}_2) \pm t* \cdot s_p \cdot \sqrt{\frac{2}{n}}\]where \(t*\) is the critical value from the t-distribution that corresponds to the desired confidence level, and \(s_p\) is the pooled standard deviation.
In our example, we calculate a 95% confidence interval, meaning we are 95% confident the true difference in means lies within the interval we computed. If the interval does not contain zero, it suggests a statistically significant difference in film speeds between the two thicknesses.
null hypothesis
The **null hypothesis** (\( H_0 \)) is a statement that there is no effect or no difference, and it serves as a starting point for hypothesis testing. It represents the idea that any observed effect in the data is due to random chance. In statistical testing, we often seek to gather evidence against the null hypothesis to support an alternative explanation.
For the film speed experiment, the null hypothesis is:\[ H_0: \mu_1 = \mu_2 \]where \(\mu_1\) and \(\mu_2\) are the true mean speeds of the 25-mil and 20-mil films respectively. The claim here is that there is no difference in the mean film speeds when thickness is altered. In our hypothesis test, we compute a test statistic and corresponding p-value to determine if there's enough evidence to reject this hypothesis.
Rejecting the null hypothesis suggests that observed differences in sample means are unlikely to be due to chance alone, implying a potential effect or difference exists.
alternative hypothesis
The **alternative hypothesis** (\( H_a \)) represents what we try to demonstrate statistically: that there is an effect or a significant difference between populations. This hypothesis is considered when the null hypothesis is rejected, indicating that the data provides sufficient evidence of a real difference.
In the context of the photoconductor film speed test, the alternative hypothesis is:\[ H_a: \mu_1 > \mu_2 \]This implies that the mean speed for the 25-mil film is greater than that for the 20-mil film (considering the decrease in microjoules per square inch corresponds to higher speed). Thus, if the alternative hypothesis is true, reducing the thickness does indeed increase the film speed significantly.
When conducting the hypothesis test, if the calculated p-value is lower than the predetermined significance level, it suggests enough evidence to accept the alternative hypothesis. This leads to the conclusion that the thinner film likely results in higher speed.

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Most popular questions from this chapter

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