/*! 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 17 The following information was ob... [FREE SOLUTION] | 91Ó°ÊÓ

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The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. \(\begin{array}{lllllllll}\text { Sample 1: } & 47.7 & 46.9 & 51.9 & 34.1 & 65.8 & 61.5 & 50.2 & 40.8 \\ & 53.1 & 46.1 & 47.9 & 45.7 & 49.0 & & & \\ \text { Sample 2: } & 50.0 & 47.4 & 32.7 & 48.8 & 54.0 & 46.3 & 42.5 & 40.8 \\ & 39.0 & 68.2 & 48.5 & 41.8 & & & & \end{array}\) a. Let \(\mu_{1}\) be the mean of population 1 and \(\mu_{2}\) be the mean of population \(2 .\) What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(98 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Test at a \(1 \%\) significance level if \(\mu_{1}\) is greater than \(\mu_{2}\).

Short Answer

Expert verified
a. The point estimate of \(\mu_{1}-\mu_{2}\) is the difference of the sample means. b. The 98% confidence interval for \(\mu_{1}-\mu_{2}\) is calculated as the difference in sample means \(\pm\) the multiplication of t critical value for 98% confidence level and standard error. c. The test conclusion is based on whether calculated t is greater than the critical t value or not.

Step by step solution

01

Calculation of Mean of Each Population

Compute the sample mean (\(\overline{x_1}, \overline{x_2}\)) for each sample. This is done by summing all numbers in a sample, then divide by the sample size.
02

Point Estimate

Calculate the point estimate of \(\mu_{1}-\mu_{2}\) by subtracting \(\overline{x_1}\) and \(\overline{x_2}\). This gives an estimated mean difference.
03

Calculation of Standard Deviation of Each Sample

Compute standard deviation (s) for each sample. This quantifies the amount of variation or dispersion of a set of values.
04

Standard Error Calculation

Calculate the standard error using the formula \(SE = \sqrt{(s_{1}^{2}/n_{1}) + (s_{2}^{2}/n_{2})}\) where \(n_{1}\) and \(n_{2}\) are the sizes of the two samples respectively
05

Confidence Interval

Using standard error (SE), the difference in sample means (DM), and the t-distribution critical value (t*) for a 98% confidence level, construct the confidence interval for \(\mu_{1}-\mu_{2}\). The confidence interval is calculated as \(DM \pm t* \times SE\)
06

Hypotheses Formulation

Formulate the null hypothesis (H0: \(\mu_{1} = \mu_{2}\)) and the alternative hypothesis (H1: \(\mu_{1} > \mu_{2}\)) to be tested at a 1% significance level.
07

T-Test Calculation

Calculate the value of the t-test statistic using the formula \(t = (DM−0)/SE\). Compare this value with the critical value of t from the t-distribution table for 1% significance level.
08

Conclusions

Based on comparison, if the calculated t is greater than the critical t value, then reject the null hypothesis and accept the alternative hypothesis. Otherwise, do not reject the null hypothesis.

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Key Concepts

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

Mean Difference
The mean difference is a central concept in statistical inference, particularly when comparing two sets of data. It helps us understand how the average values of two populations (or samples) differ.

To find the mean difference between two samples, first compute the mean of each sample. This involves averaging the data within each sample. Once these averages, or means, are computed, the mean difference is simply the subtraction of the two sample means, which can be represented mathematically as \[\overline{x_1} - \overline{x_2}\]where \(\overline{x_1}\) and \(\overline{x_2}\) are the means of sample 1 and sample 2, respectively.

Understanding the mean difference leads to insights into whether the observed difference is significant, or just a result of random variation. This can have important implications in fields like medicine, science, and social sciences, where comparing means is frequently used to infer interesting insights. Always remember, the mean difference by itself does not consider variability; hence, further statistical tests are usually needed for making strong inferences.
Confidence Interval
A confidence interval provides a range of values, derived from the sample data, that is likely to contain the value of an unknown population parameter. It gives an interval estimate of the potential difference between the population means (\( \mu_1 - \mu_2 \)).

To construct a confidence interval, you'll need a few key pieces of information:
  • The mean difference between the samples (already calculated previously).
  • The standard error, which takes into account how much the sample means might vary.
  • A critical value from the statistical distribution, which depends on the desired confidence level (e.g., 98% confidence level).
With these, the confidence interval for the mean difference is computed as \[ DM \pm t* \times SE\]where \(DM\) is the difference in means, \(t*\) is the critical t value from the t-distribution table, and \(SE\) is the standard error. The confidence interval thus gives a range where we expect the true mean difference to fall.

It's crucial to understand that the wider the interval, the less precise our estimate, but higher the likelihood (confidence) that the interval includes the true mean difference. Always remember, the confidence interval does not imply probability about individual samples but rather describes the uncertainty about our parameter estimates.
Hypothesis Testing
Hypothesis testing is a statistical method that uses sample data to evaluate a hypothesis about a population parameter, in this case, the difference in means \(\mu_1\) and \(\mu_2\).

The process begins with the formulation of two hypotheses:
  • The null hypothesis \(H_0\): States there is no difference, or the difference is zero, \(\mu_1 = \mu_2\).
  • The alternative hypothesis \(H_1\): States \(\mu_1 > \mu_2\), suggesting that population 1 has a larger mean.
After we state our hypotheses, we perform a t-test to calculate a t-statistic.This is done using the formula:\[t = \frac{DM - 0}{SE}\]where \(DM\) is the difference in sample means and \(SE\) is the standard error. Comparing this t-statistic to a critical t value from the t-distribution table allows us to make an inference. If the calculated t is larger than the critical t value at the chosen significance level (like 1%), we reject \(H_0\).

The significance level, such as 1%, is the probability threshold for deciding if our observed sample result is sufficiently extreme to reject \(H_0\). Performing hypothesis testing in statistical analysis allows one to ascertain whether sample data can support a specific claim about the population, beyond what might be expected by chance alone.

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Most popular questions from this chapter

We wish to estimate the difference between the mean scores on a standardized test of students taught by Instructors \(\mathrm{A}\) and \(\mathrm{B}\). The scores of all students taught by Instructor A have a normal distribution with a standard deviation of 15, and the scores of all students taught by Instructor \(\mathrm{B}\) have a normal distribution with a standard deviation of \(10 .\) To estimate the difference between the two means, you decide that the same number of students from each instructor's class should be observed. a. Assuming that the sample size is the same for each instructor's class, how large a sample should be taken from each class to estimate the difference between the mean scores of the two populations to within 5 points with \(90 \%\) confidence? b. Suppose that samples of the size computed in part a will be selected in order to test for the difference between the two population mean scores using a .05 level of significance. How large does the difference between the two sample means have to be for you to conclude that the two population means are different? c. Explain why a paired-samples design would be inappropriate for comparing the scores of Instructor A versus Instructor \(\mathrm{B}\).

Gamma Corporation is considering the installation of governors on cars driven by its sales staff. These devices would limit the car speeds to a preset level, which is expected to improve fuel economy. The company is planning to test several cars for fuel consumption without governors for 1 week. Then governors would be installed in the same cars, and fuel consumption will be monitored for another week. Gamma Corporation wants to estimate the mean difference in fuel consumption with a margin of error of estimate of \(2 \mathrm{mpg}\) with a \(90 \%\) confidence level. Assume that the differences in fuel consumption are normally distributed and that previous studies suggest that an estimate of \(s_{d}=3 \mathrm{mpg}\) is reasonable. How many cars should be tested? (Note that the critical value of \(t\) will depend on \(n\), so it will be necessary to use trial and error.)

Manufacturers of two competing automobile models, Gofer and Diplomat, each claim to have the lowest mean fuel consumption. Let \(\mu_{1}\) be the mean fuel consumption in miles per gallon (mpg) for the Gofer and \(\mu_{2}\) the mean fuel consumption in mpg for the Diplomat. The two manufacturers have agreed to a test in which several cars of each model will be driven on a 100 -mile test run. The fuel consumption, in \(\mathrm{mpg}\), will be calculated for each test run. The average mpg for all 100 -mile test runs for each model gives the corresponding mean. Assume that for each model the gas mileages for the test runs are normally distributed with \(\sigma=2 \mathrm{mpg}\). Note that each car is driven for one and only one 100 -mile test run. a. How many cars (i.e., sample size) for each model are required to estimate \(\mu_{1}-\mu_{2}\) with a \(90 \%\) confidence level and with a margin of error of estimate of \(1.5 \mathrm{mpg}\) ? Use the same number of cars (i.e., sample size) for each model. b. If \(\mu_{1}\) is actually \(33 \mathrm{mpg}\) and \(\mu_{2}\) is actually \(30 \mathrm{mpg}\), what is the probability that five cars for each model would yield \(\bar{x}_{1} \geq \bar{x}_{2}\) ?

In a survey of American drivers, \(79 \%\) of women drivers and \(85 \%\) of men drivers said that they exceeded the speed limit at least once in the past week. Suppose that these percentages are based on random samples of 600 women and 700 men drivers. a. Let \(p_{1}\) and \(p_{2}\) be the proportion of all women and men American drivers, respectively, who will say that they exceeded the speed limit at least once in the past week. Construct a \(98 \%\) confidence interval for \(p_{1}-p_{2}\). b. Using a \(1 \%\) significance level, can you conclude that \(p_{1}\) is lower than \(p_{2}\) ? Use both the critical-value and the \(p\) -value approaches.

The manager of a factory has devised a detailed plan for evacuating the building as quickly as possible in the event of a fire or other emergency. An industrial psychologist believes that workers actually leave the factory faster at closing time without following any system. The company holds fire drills periodically in which a bell sounds and workers leave the building according to the system. The evacuation time for each drill is recorded. For comparison, the psychologist also records the evacuation time when the bell sounds for closing time each day. A random sample of 36 fire drills showed a mean evacuation time of \(5.1\) minutes with a standard deviation of \(1.1\) minutes. \(\mathrm{A}\)random sample of 37 days at closing time showed a mean evacuation time of \(4.2\) minutes with a standard deviation of \(1.0\) minute. a. Construct a \(99 \%\) confidence interval for the difference between the two population means. b. Test at a \(5 \%\) significance level whether the mean evacuation time is smaller at closing time than during fire drills. Assume that the evacuation times at closing time and during fire drills have unknown and unequal population standard deviations.

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