91Ó°ÊÓ

For Exercises 9 through \(24,\) 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. Museum Attendance A metropolitan children's museum open year-round wants to see if the variance in daily attendance differs between the summer and winter months. Random samples of 30 days each were selected and showed that in the winter months, the sample mean daily attendance was 300 with a standard deviation of \(52,\) and the sample mean daily attendance for the summer months was 280 with a standard deviation of \(65 .\) At \(\alpha=0.05,\) can we conclude a difference in variances?

Step by step solution

01

State the Hypotheses

To test the variance difference, we set up the null and alternative hypotheses. Let \( \sigma_1^2 \) be the variance for winter and \( \sigma_2^2 \) for summer. The null hypothesis \( H_0 \) is \( \sigma_1^2 = \sigma_2^2 \), meaning there is no difference in variances. The alternative hypothesis \( H_a \) is \( \sigma_1^2 eq \sigma_2^2 \), meaning there is a difference in variances. The claim is the alternative hypothesis, that there is a difference in variances.

02

Find the Critical Value

The test statistic is the F-distribution because we are comparing two variances. We calculate the critical value for a two-tailed test with \( n_1 - 1 = 29 \) and \( n_2 - 1 = 29 \) degrees of freedom at \( \alpha = 0.05 \). Using F-distribution tables or a calculator, the critical values are \( F_{0.025,29,29} \approx 1.88 \) for the lower bound and \( F_{0.975,29,29} \approx 0.532 \) for the upper bound.

03

Compute the Test Value

We calculate the F-test statistic using the sample variances. The test statistic is \( F = \frac{s_1^2}{s_2^2} = \frac{52^2}{65^2} = \frac{2704}{4225} \approx 0.639 \).

04

Make the Decision

Compare the test statistic \( F = 0.639 \) with the critical values \( 1.88 \) and \( 0.532 \). Since \( 0.639 \) is not in the critical region (not less than 0.532 or greater than 1.88), we do not reject the null hypothesis.

05

Summarize the Results

At the \( \alpha = 0.05 \) significance level, we do not have sufficient evidence to claim that there is a difference in variances of daily attendance between winter and summer months at the museum. Thus, the variance in attendance is not significantly different between the two seasons.

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Key Concepts

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

F-distribution

The F-distribution is a key concept when you're comparing variances. It is used primarily to understand and analyze data variances between two populations. When we speak of variances, we are looking at how much the data points deviate or spread out from the mean value. The F-distribution is always skewed to the right and depends on both the numerator and the denominator degrees of freedom. For the exercise at hand, the numerators and denominators both have 29 degrees of freedom, as determined by the sample size minus one.

• It is named after the statistician R.A. Fisher.
• The F-distribution has a continuous probability distribution.
• A characteristic of this distribution is that it's always non-negative, as it's based on squared variances.

In hypothesis testing, the F-distribution helps us decide whether to reject the null hypothesis by comparing it with critical F-values from a table or statistical software.

Null Hypothesis

The null hypothesis is a foundational concept in hypothesis testing. It posits that there is no significant difference or effect. In this exercise, the null hypothesis ( H_0 ) was that the variances of daily museum attendance between winter and summer are equal. This assumption works as our default position that we either reject or fail to reject based on our statistical analysis.

• Represents no change or status quo situation.
• It is generally tested against the alternative hypothesis.
• Failure to reject the null hypothesis implies that any differences observed are due to chance or natural variability.

The null hypothesis serves as a benchmark to measure any deviation —a critical aspect of many scientific and statistical inquiries.

Alternative Hypothesis

Opposite to the null hypothesis, the alternative hypothesis is what researchers often aim to prove. In this case, we're suggesting that there is a difference in variances between summer and winter museum attendance rates. This alternative hypothesis ( H_a ) underlines the need for further analysis to see if the hypothesis holds true.

• It suggests the presence of an effect or a significant difference.
• Usually represents the hypothesis of interest.
• A test statistic that falls in the critical region leads to the rejection of the null hypothesis, thereby supporting the alternative hypothesis.

When conducting a hypothesis test, a significant result would mean accepting the alternative hypothesis, indicating a true effect rather than random chance.

Variance Comparison

In statistics, comparing variances is crucial to understanding the spread of different datasets. Variance comparison helps in determining how much individual data elements deviate from the mean. In our exercise, we're analyzing whether the variance of museum attendance on different days in summer differs from winter. This requires calculating the sample variances and then comparing them using the F-test.

• Comparison can help identify periods of stability or variability in a dataset.
• It is a standard practice to check for homogeneity of variances before proceeding to other tests, like ANOVA.
• Assists in determining the robustness and reliability of data.

In this scenario, calculating the test statistic using the given standard deviations was necessary to draw conclusions about any significant variance differences between two populations.