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Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A researcher wishes to see if the variance of the areas in square miles for counties in Indiana is less than the variance of the areas for counties in Iowa. A random sample of counties is selected, and the data are shown. At \(\alpha=0.01,\) can it be concluded that the variance of the areas for counties in Indiana is less than the variance of the areas for counties in Iowa? $$ \begin{array}{llll|llll} &&&{\text { Indiana }} &{\text { Iowa }} \\ \hline 406 & 393 & 396 & 485 & 640 & 580 & 431 & 416 \\ 431 & 430 & 369 & 408 & 443 & 569 & 779 & 381 \\ 305 & 215 & 489 & 293 & 717 & 568 & 714 & 731 \\ 373 & 148 & 306 & 509 & 571 & 577 & 503 & 501 \\ 560 & 384 & 320 & 407 & 568 & 434 & 615 & 402 \end{array} $$

Short Answer

Expert verified
At α = 0.01 , determine if F_{calculated} < F_{critical} to support the claim that Indiana counties have less variance.

Step by step solution

01

State Hypotheses

We need to determine if there is a difference in the variances of the areas of counties between Indiana and Iowa. Therefore, we define our null and alternative hypotheses as follows: - Null hypothesis ( H_0 ): The variance of the areas of counties in Indiana ( σ^2_{IN} ) is equal to the variance of the areas of counties in Iowa ( σ^2_{IA} ) i.e., σ^2_{IN} = σ^2_{IA} . - Alternative hypothesis ( H_1 ): The variance of the areas of counties in Indiana is less than the variance of the areas of counties in Iowa i.e., σ^2_{IN} < σ^2_{IA} . The claim in this problem is the alternative hypothesis ( H_1 ).
02

Find the Critical Value

Since we are testing for variance, we use the F-distribution and perform an F-test. To find the critical value, we use an α level of 0.01. 1. Calculate the degrees of freedom for each sample. For Indiana: df_{IN} = n_{IN} - 1 = 20 - 1 = 19 . For Iowa: df_{IA} = n_{IA} - 1 = 16 - 1 = 15 . 2. Use the F-distribution table to find the critical F-value for df_1 = 15 and df_2 = 19 . This value at α = 0.01 for a one-tailed test is approximately F_{critical} ≈ 0.351 .
03

Compute the Test Value

Calculate the sample variances for each dataset and then the test statistic:1. Find the sample variance for Indiana (s^2_{IN}) and Iowa (s^2_{IA}).2. Using the formula:\[ F = \frac{s^2_{IA}}{s^2_{IN}} \]3. Calculate s^2_{IN} and s^2_{IA} from data, and substitute into the formula to find F.
04

Make the Decision

Compare the calculated F-value to the critical value from the F-distribution table. If F_{calculated} < F_{critical} , then reject the null hypothesis. If F_{calculated} ≥ F_{critical} , then do not reject the null hypothesis. This determines whether the variance for Indiana counties is indeed less than that for Iowa.
05

Summarize Results

After the comparison, conclude the following: - If you rejected the null hypothesis, the researcher's claim is supported: that Indiana counties have less variance in area compared to Iowa. - If you did not reject the null hypothesis, there is insufficient evidence to support the claim.

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Key Concepts

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

Variance
Variance is a statistical measure that represents the dispersion or spread of a set of data points. It shows how much the numbers in a dataset differ from the mean, or average, value. In essence, variance helps us understand how spread out the data is. A high variance indicates that the numbers are spread out over a wider range, while a low variance means they are closer to the mean.
To calculate variance, follow these steps:
  • Find the mean (average) of the dataset.
  • Subtract the mean from each number to find the deviation of each number from the mean.
  • Square each of these deviations to eliminate negative values.
  • Calculate the average of these squared deviations. This average is the variance.
The formula for variance (σ^2) is:\[σ^2 = \frac{\sum{(x_i - \mu)^2}}{n}\]Where \(x_i\) is each data point, \(\mu\) is the mean, and \(n\) is the number of data points.
Normal Distribution
Normal distribution, often known as a bell curve, describes how the values of a variable are distributed. It is called 'normal' for a good reason: many phenomena in the natural world approximate this distribution. The graph of a normal distribution is symmetrical, with most of the observations clustering around the central peak and fewer observations farther from the center.
Key characteristics of a normal distribution include:
  • Symmetrical shape: Its left and right sides are mirror images.
  • Mean, median, and mode are all equal and located at the center of the distribution.
  • A specific pattern of data spread such that approximately 68% of the data lies within one standard deviation, 95% within two, and 99.7% within three standard deviations.
Normal distributions are critical in hypothesis testing as they allow the use of a plethora of statistical tools. When a dataset follows a normal distribution, it assumes that most data points fall around the average, making predictions and decisions based on this model quite reliable.
F-distribution
The F-distribution is primarily used in hypothesis tests involving variances. It is the distribution of the ratio of two sample variances and is essential in performing variance analysis, like the F-test. The F-distribution is skewed and depends on two types of degrees of freedom: numerator and denominator degrees of freedom.
Here's how it is utilized in hypothesis testing:
  • Numerator degrees of freedom come from the number of observations from which one sample variance is calculated.
  • Denominator degrees of freedom are derived from the number of observations from which the other sample variance is computed.
  • Using these degrees, the F-table provides critical values. If the calculated F-value from your data is less than the critical F-value, the null hypothesis can be rejected.
Its skewness means there is a different value for each pair of numerator and denominator degrees of freedom. In hypothesis testing for variances, the F-distribution allows us to compare if the observed variance differences between two samples are significant or not.
Statistical Hypotheses
Statistical hypotheses are assumptions about a population parameter that we test using statistical methods. In hypothesis testing, we begin with two opposing hypotheses: the null hypothesis and the alternative hypothesis.
Here's how they function:
  • Null hypothesis (H_0): It is the statement that there is no effect or difference. It is assumed true until evidence suggests otherwise.
  • Alternative hypothesis (H_1): This is what you want to prove. It states that there is a genuine effect or difference.
After establishing the hypotheses, data is used to perform tests that will lead to a decision about the hypotheses. The outcome helps in deciding whether to reject the null hypothesis in favor of the alternative or not to reject it. This process is fundamental in research as it helps in making informed decisions based on statistical evidence.

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Most popular questions from this chapter

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. The overall U.S. public high school graduation rate is \(73.4 \% .\) For Pennsylvania it is \(83.5 \%\) and for Idaho \(80.5 \%-\) a difference of \(3 \% .\) Random samples of 1200 students from each state indicated that 980 graduated in Pennsylvania and 940 graduated in Idaho. At the 0.05 level of significance, can it be concluded that there is a difference in the proportions of graduating students between the states?

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A researcher wishes to see if there is a difference in the manual dexterity of athletes and that of band members. Two random samples of 30 are selected from each group and are given a manual dexterity test. The mean of the athletes' test was \(87,\) and the mean of the band members' test was \(92 .\) The population standard deviation for the test is \(7.2 .\) At \(\alpha=0.01,\) is there a significant difference in the mean scores?

Perform each of these steps. Assume that all variables are normally or approximately normally distributed a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Overweight Dogs A veterinary nutritionist developed a diet for overweight dogs. The total volume of food consumed remains the same, but one-half of the dog food is replaced with a low-calorie "filler" such as canned green beans. Six overweight dogs were randomly selected from her practice and were put on this program. Their initial weights were recorded, and they were weighed again after 4 weeks. At the 0.05 level of significance, can it be concluded that the dogs lost weight? $$ \begin{array}{l|llllll} \text { Before } & 42 & 53 & 48 & 65 & 40 & 52 \\ \hline \text { After } & 39 & 45 & 40 & 58 & 42 & 47 \end{array} $$

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. The average monthly Social Security benefit for a specific year for retired workers was \(\$ 954.90\) and for disabled workers was \(\$ 894.10 .\) Researchers used data from the Social Security records to test the claim that the difference in monthly benefits between the two groups was greater than \(\$ 30 .\) Based on the following information, can the researchers' claim be supported at the 0.05 level of significance? $$ \begin{array}{lll} & \text { Retired } & \text { Disabled } \\ \hline \text { Sample size } & 60 & 60 \\ \text { Mean benefit } & \$ 960.50 & \$ 902.89 \\ \text { Population standard deviation } & \$ 98 & \$ 101 \end{array} $$

What are the two different degrees of freedom associated with the \(F\) distribution?

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