/*! 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 1 The following results come from ... [FREE SOLUTION] | 91Ó°ÊÓ

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The following results come from two independent random samples taken of two populations. Sample 1 \(\quad\) Sample 2 \(\begin{array}{ll}n_{1}=50 & n_{2}=35 \\ \bar{x}_{1}=13.6 & \bar{x}_{2}=11.6 \\\ \sigma_{1}=2.2 & \sigma_{2}=3.0\end{array}\) a. What is the point estimate of the difference between the two population means? b. Provide a \(90 \%\) confidence interval for the difference between the two population means. c. Provide a \(95 \%\) confidence interval for the difference between the two population means.

Short Answer

Expert verified
a) 2.0; b) (1.021, 2.979); c) (0.834, 3.166)

Step by step solution

01

Identify the Point Estimate

To find the point estimate of the difference between the two population means, subtract the sample mean of Population 2 from the sample mean of Population 1. The formula is: \[ \text{Point estimate} = \bar{x}_1 - \bar{x}_2 \]Substitute the given values: \[ \bar{x}_1 = 13.6, \quad \bar{x}_2 = 11.6 \]So, the point estimate is:\[ 13.6 - 11.6 = 2.0 \]
02

Calculate Standard Error for the Difference in Means

The standard error of the difference between two independent sample means is calculated using the formula:\[ SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \]where \( \sigma_1 = 2.2 \), \( n_1 = 50 \), \( \sigma_2 = 3.0 \), and \( n_2 = 35 \).Calculate:\[ SE = \sqrt{\frac{2.2^2}{50} + \frac{3.0^2}{35}} = \sqrt{\frac{4.84}{50} + \frac{9.0}{35}} \approx \sqrt{0.0968 + 0.2571} \approx \sqrt{0.3539} \approx 0.595 \]
03

Confidence Interval for 90%

To calculate a 90% confidence interval, use the formula:\[ (\text{Point estimate}) \pm z_{0.05} \times SE \]For a 90% confidence level, the z-value is approximately 1.645.Using the point estimate from Step 1 (2.0) and the standard error from Step 2 (0.595):\[ 2.0 \pm 1.645 \times 0.595 \approx 2.0 \pm 0.979 \]Thus, the 90% confidence interval is: \[ (1.021, 2.979) \]
04

Confidence Interval for 95%

For a 95% confidence interval, use:\[ (\text{Point estimate}) \pm z_{0.025} \times SE \]The z-value for a 95% confidence level is approximately 1.96.So:\[ 2.0 \pm 1.96 \times 0.595 \approx 2.0 \pm 1.166 \]Thus, the 95% confidence interval is: \[ (0.834, 3.166) \]

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Key Concepts

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

Point Estimate
A point estimate is a single value that serves as a best guess or a representative value for a statistical parameter, like the population mean difference in this context. It is calculated directly from sample data and provides a quick overview of what the actual parameter might be.

For instance, in the exercise, we want to find out the possible difference between two population means. We use the sample means from each population. The formula is:
  • \( \text{Point estimate} = \bar{x}_1 - \bar{x}_2 \)
Using this, we subtract the mean of Sample 2 (\( \bar{x}_2 = 11.6 \)) from the mean of Sample 1 (\( \bar{x}_1 = 13.6 \)). Thus, the point estimate is:
  • \( 13.6 - 11.6 = 2.0 \)
This simple subtraction gives us a specific number, 2.0, suggesting how much we believe one population mean exceeds the other based on our samples.
Standard Error
The standard error measures the variance of a sampling distribution. Essentially, it tells us how much variability there is in our point estimates when computed over numerous samples. This is crucial for understanding the accuracy and reliability of our statistical estimates.

To compute the standard error of the difference between independent sample means, we use:
  • \[ SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \]
where:
  • \( \sigma_1 \) and \( \sigma_2 \) are standard deviations of each sample population, and
  • \( n_1 \) and \( n_2 \) are the sample sizes.
Applying values from the exercise:
  • \( \sigma_1 = 2.2, n_1 = 50, \sigma_2 = 3.0, n_2 = 35 \)
The calculated standard error is approximately 0.595. This number indicates the standard distribution of sample means' differences if we repeatedly took samples.
Independent Samples
Independent samples refer to two separate groups that do not influence each other, meaning the outcomes from one group do not affect the outcomes from another. This is an important assumption in many statistical tests, as it justifies the application of certain statistical formulas without accounting for potential biases.

In our context, because the samples are drawn independently, the mean and variance calculations assume that each group is unaffected by the other.

This independent assumption is crucial when conducting analysis aimed at comparing these group characteristics.
  • It allows the use of the formulas for point estimates and standard errors before confidently calculating the confidence interval.
This property ensures that the data we analyze are both unbiased and representative, as it relies on the lack of relationship between the two sample groups.
Population Means
Population means refer to the average values within a complete group that we are interested in understanding or making conclusions about. In practical terms, these are often unknown because it's rarely feasible to collect data on an entire population. Instead, we rely on sample data to make educated guesses about these means.

When we estimate population means using sample data, the sample mean serves as a representation of the larger population's mean. Such estimations rely heavily on consistent and unbiased data collection methods. Combined with statistical tools like point estimates and confidence intervals, we can gauge and understand the potential range of these population means.
  • For instance, any variability in a population mean typically comes from the differences in sample means we collect and analyze.
  • Population mean inference becomes more robust with larger sample sizes and diverse representation.
Having a clear grasp of how these sample estimates relate to the actual population means helps us create accurate models and predictions of what's truly happening within a broader scope.

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Most popular questions from this chapter

During the 2003 season, Major League Baseball took steps to speed up the play of baseball games in order to maintain fan interest (CNN Headline News, September 30, 2003). The following results come from a sample of 60 games played during the summer of 2002 and a sample of 50 games played during the summer of \(2003 .\) The sample mean shows the mean duration of the games included in each sample. \(\begin{array}{cl}\text { 2002 Season } & \text { 2003 Season } \\ n_{1}=60 & n_{2}=50 \\ \bar{x}_{1}=2 \text { hours, } 52 \text { minutes } & \bar{x}_{2}=2 \text { hours, } 46 \text { minutes }\end{array}\) a. \(\quad\) A research hypothesis was that the steps taken during the 2003 season would reduce the population mean duration of baseball games. Formulate the null and alternative hypotheses. b. What is the point estimate of the reduction in the mean duration of games during the 2003 season? c. Historical data indicate a population standard deviation of 12 minutes is a reasonable assumption for both years. Conduct the hypothesis test and report the \(p\) -value. At a .05 level of significance, what is your conclusion? d. Provide a \(95 \%\) confidence interval estimate of the reduction in the mean duration of games during the 2003 season. e. What was the percentage reduction in the mean time of baseball games during the 2003 season? Should management be pleased with the results of the statistical analysis? Discuss. Should the length of baseball games continue to be an issue in future years? Explain.

Airline travelers often choose which airport to fly from based on flight cost. cost data (in dollars) for a sample of flights to eight cities from Dayton, Ohio, and Louisville, Kentucky, were collected to help determine which of the two airports was more costly to fly from (The Cincinnati Enquirer; February 19,2006 ). A researcher argued that it is significantly more costly to fly out of Dayton than Louisville. Use the sample data to see whether they support the researcher's argument. Use \(\alpha=.05\) as the level of significance. $$\begin{array}{lcr} \text { Destination } & \text { Dayton } & \text { Louisville } \\ \text { Chicago-O'Hare } & \$ 319 & \$ 142 \\ \text { Grand Rapids, Michigan } & 192 & 213 \\ \text { Portland, Oregon } & 503 & 317 \\ \text { Atlanta } & 256 & 387 \\ \text { Seattle } & 339 & 317 \\ \text { South Bend, Indiana } & 379 & 167 \\ \text { Miami } & 268 & 273 \\ \text { Dallas-Ft. Worth } & 288 & 274 \end{array}$$

Consider the following data for two independent random samples taken from two normal populations. $$\begin{array}{l|rrrrrr} \text { Sample 1 } & 10 & 7 & 13 & 7 & 9 & 8 \\ \hline \text { Sample 2 } & 8 & 7 & 8 & 4 & 6 & 9 \end{array}$$ a. Compute the two sample means. b. Compute the two sample standard deviations. c. What is the point estimate of the difference between the two population means? d. What is the \(90 \%\) confidence interval estimate of the difference between the two population means?

Typical prices of single-family homes in the state of Florida are shown for a sample of 15 metropolitan areas (Naples Daily News, February 23, 2003). Data are in thousands of dollars. $$\begin{array}{lcr} \text { Metropolitan Area } & \text { January } 2003 & \text { January } 2002 \\\ \text { Daytona Beach } & 117 & 96 \\ \text { Fort Lauderdale } & 207 & 169 \\ \text { Fort Myers } & 143 & 129 \\ \text { Fort Walton Beach } & 139 & 134 \\ \text { Gainesville } & 131 & 119 \\ \text { Jacksonville } & 128 & 119 \\ \text { Lakeland } & 91 & 85 \\ \text { Miami } & 193 & 165 \\ \text { Naples } & 263 & 233 \\ \text { Ocala } & 86 & 90 \\ \text { Orlando } & 134 & 121 \\ \text { Pensacola } & 111 & 105 \\ \text { Sarasota-Bradenton } & 168 & 141 \\ \text { Tallahassee } & 140 & 130 \\ \text { Tampa-St. Petersburg } & 139 & 129 \end{array}$$ a. Use a matched-sample analysis to develop a point estimate of the population mean one-year increase in the price of single-family homes in Florida. b. Develop a \(90 \%\) confidence interval estimate of the population mean one- year increase in the price of single-family homes in Florida. c. What was the percentage increase over the one-year period?

Consider the following hypothesis test. $$\begin{array}{l} H_{0}: \mu_{d} \leq 0 \\ H_{\mathrm{a}}: \mu_{d}>0 \end{array}$$ The following data are from matched samples taken from two populations. $$\begin{array}{ccc} & {\text { Population }} \\ \text { Element } & \mathbf{1} & \mathbf{2} \\ 1 & 21 & 20 \\ 2 & 28 & 26 \\ 3 & 18 & 18 \\ 4 & 20 & 20 \\ 5 & 26 & 24 \end{array}$$ a. Compute the difference value for each element. b. Compute \(\bar{d}\) c. Compute the standard deviation \(s_{d}\) d. Conduct a hypothesis test using \(\alpha=.05 .\) What is your conclusion?

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