91Ó°ÊÓ

A factory that emits airborne pollutants is testing two different brands of filters for its smokestacks. The factory has two smokestacks. One brand of filter (Filter I) is placed on one smokestack, and the other brand (Filter II) is placed on the second smokestack. Random samples of air released from the smokestacks are taken at different times throughout the day. Pollutant concentrations are measured from both stacks at the same time. The following data represent the pollutant concentrations (in parts per million) for samples taken at 20 different times after passing through the filters. Assume that the differences in concentration levels at all times are approximately normally distributed. $$ \begin{array}{cccccc} \hline \text { Time } & \text { Filter I } & \text { Filter II } & \text { Time } & \text { Filter I } & \text { Filter II } \\ \hline 1 & 24 & 26 & 11 & 11 & 9 \\ 2 & 31 & 30 & 12 & 8 & 10 \\ 3 & 35 & 33 & 13 & 14 & 17 \\ 4 & 32 & 28 & 14 & 17 & 16 \\ 5 & 25 & 23 & 15 & 19 & 16 \\ 6 & 25 & 28 & 16 & 19 & 18 \\ 7 & 29 & 24 & 17 & 25 & 27 \\ 8 & 30 & 33 & 18 & 20 & 22 \\ 9 & 26 & 22 & 19 & 23 & 27 \\ 10 & 18 & 18 & 20 & 32 & 31 \\ \hline \end{array} $$ a. Make a \(95 \%\) confidence interval for the mean of the population paired differences, where a paired difference is equal to the pollutant concentration passing through Filter I minus the pollutant concentration passing through Filter II. b. Using the \(5 \%\) significance level, can you conclude that the average paired difference for concentration levels is different from zero?

Step by step solution

01

Compile the Paired Differences

Calculate the paired differences for each time by subtracting the concentration of Filter II from Filter I. Let's denote these differences as \(d\). Then, calculate the mean paired difference \(\bar{d}\) and the standard deviation \(s_d\) of these differences.

02

Formulate the Hypotheses

The null hypothesis \(H_0\) is that the mean difference in the population equals 0, i.e, \(µ_{d} = 0\). The alternative hypothesis \(H_1\) is that the mean difference in the population is not equal to 0, i.e, \(µ_{d} ≠ 0\).

03

Calculate the Test Statistic

The paired t statistic is calculated using the formula: \(t = \frac{\bar{d}}{s_d/\sqrt{n}}\), where \(n\) is number of paired observations. Compare the calculated \(t\) value with the critical \(t\) value from t table at \( \alpha/2 = 0.025\) for \( n-1 \) degrees of freedom.

04

Construct the Confidence Interval

The 95% confidence interval for the mean difference is given by the formula: \(\bar{d} \pm t_{\alpha/2} * \frac {s_d}{ \sqrt{n}}\). Here, \(t_{\alpha/2}\) is the critical value from the t-distribution table for the desired level of confidence, 95%, and \( \alpha/2 = 0.025 \).

05

Determine if the hypothesis can be rejected

Compare the calculated \(t\) value with the critical \(t\) value. If the calculated \(t\) value falls in the rejection region (either greater than the positive critical \(t\) value or smaller than the negative critical \(t\) value), we reject the null hypothesis.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Confidence Interval

When we talk about a confidence interval, we are referring to a range of values that is likely to contain an unknown population parameter. In this exercise, we are interested in the mean of the population paired differences in pollutant concentrations after using two different filters.
This gives us a 95% confidence interval, which means that we can be 95% sure that the true mean difference falls within this range.
For our paired differences, the confidence interval is calculated as:

• Take the mean of all the differences between Filter I and Filter II.
• Add and subtract the product of the critical t-value and standard error to/from the mean.

This interval provides insight into whether Filter I or Filter II reduces pollution more effectively.

Null Hypothesis

The null hypothesis, often denoted by \( H_0 \), is a statement of no effect or no difference.
In our scenario, the null hypothesis asserts that there is no difference in pollutant reduction between the two filters, i.e., the mean of the paired differences equals zero:

• \( \mu_d = 0 \) - where \( \mu_d \) is the population mean of the paired differences.

Testing this hypothesis helps us determine if any observed differences in pollutant levels are due to chance.

Alternative Hypothesis

The alternative hypothesis, denoted by \( H_1 \), suggests that there is an effect or a difference.
In this case, it proposes that the two filters do not perform identically, and the mean of paired differences is not zero:

• \( \mu_d eq 0 \)

This hypothesis challenges the null hypothesis and is what we attempt to support by conducting the paired t-test. If we find strong evidence against the null, it indirectly supports the alternative hypothesis.

Significance Level

The significance level, often given as \( \alpha \), is the probability of rejecting the null hypothesis when it is true. In simpler terms, it's the chance of making a type I error.

• For this exercise, the significance level is set at 5% or 0.05.

This threshold is used to determine the critical value from the t-distribution table. If the results of our test statistic fall outside of this interval, we consider that finding statistically significant, indicating a real difference between the filters.

Test Statistic

The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It helps determine whether to reject the null hypothesis.
For a paired t-test, the test statistic \( t \) is given by:\[ t = \frac{\bar{d}}{s_d/\sqrt{n}} \]Where:

• \( \bar{d} \) is the mean of paired differences.
• \( s_d \) is the standard deviation of paired differences.
• \( n \) is the number of paired observations.

By comparing this \( t \)-value to a critical value from the t-distribution, we determine whether to accept or reject the null hypothesis. If this \( t \)-value falls into the critical region defined by our significance level, it provides evidence to reject \( H_0 \) and suggests a significant difference in filter performance.