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Chapter 8: Problem 11

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. A researcher claims that the standard deviation of the number of deaths annually from tornadoes in the United States is less than \(35 .\) If a random sample of 11 years had a standard deviation of \(32,\) is the claim believable? Use \(\alpha=0.05\)

Short Answer

Expert verified
The claim is not believable based on the hypothesis test.

Step by step solution

01

State the hypotheses

First, we need to state the null and alternative hypotheses. The claim is that the standard deviation is less than 35, so we formulate the hypotheses as follows: - Null hypothesis \[ H_0: \sigma = 35 \]- Alternative hypothesis \[ H_a: \sigma < 35 \]
02

Determine the critical value

Since we are dealing with a small sample size (11 years), we will use the chi-square distribution. The degrees of freedom (df) is calculated as follows: - \( df = n - 1 = 11 - 1 = 10 \) Next, using the chi-square distribution table or calculator, find the critical value for \( \alpha = 0.05 \) in the left tail for \( df = 10 \). The critical value is approximately \( \chi^2_{0.05, 10} = 3.940 \).
03

Calculate the test statistic

The formula for the test statistic using the chi-square distribution is: \[ \chi^2 = \frac{(n-1)s^2}{\sigma_0^2} \]where \( n = 11 \), \( s = 32 \), and \( \sigma_0 = 35 \). Substituting these values into the formula gives: \[ \chi^2 = \frac{(11-1) \times 32^2}{35^2} = \frac{10 \times 1024}{1225} \approx 8.365 \]
04

Make the decision

We compare the calculated test statistic \( \chi^2 = 8.365 \) with the critical value \( \chi^2_{critical} = 3.940 \). Since \( 8.365 \) is greater than \( 3.940 \), the test statistic does not fall in the critical region. Therefore, we fail to reject the null hypothesis.
05

Conclusion

Since we failed to reject the null hypothesis, there is not enough statistical evidence to support the researcher's claim that the standard deviation of the number of deaths annually from tornadoes in the U.S. is less than 35.

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Key Concepts

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

Chi-Square Distribution
The chi-square distribution is a vital tool in statistics, especially useful for hypothesis testing about population variances. This kind of distribution is not symmetric and it shifts its shape depending on the degrees of freedom. The smaller the degrees of freedom, the more skewed the distribution appears. In hypothesis testing, this distribution helps determine if a sample variance significantly differs from a hypothesized population variance.

In the context of our exercise, the chi-square distribution is used because we are assessing the variance or standard deviation, not the mean. When the sample size is small, and the variance is unknown, the chi-square distribution becomes essential in determining the critical value to make statistical decisions.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It gives insight into the overall spread of the data points in a dataset. A low standard deviation means that most data points are close to the mean, whereas a high standard deviation indicates that the data points are spread out over a wide range of values.

In the exercise, the researcher's claim is focused on whether the standard deviation of tornado-related deaths is less than a specific value, which is 35. The standard deviation of our sample was 32, which suggests variability is somewhat less than proposed. However, statistical testing is required to ascertain if this observed difference is significant or just a result of random chance.
Critical Value
In hypothesis testing, the critical value is a threshold that determines the boundary of the rejection region. It helps in deciding whether to reject the null hypothesis. This value is influenced by the chosen significance level, commonly denoted by \( \alpha \), and the degrees of freedom.

For our problem, the critical value is derived from the chi-square distribution for a significance level of 0.05, as we're testing in the lower tail with 10 degrees of freedom. This critical threshold is found to be approximately 3.940. Comparing it to our test statistic helps determine if we should reject the claim that the standard deviation is equal to 35.
Degrees of Freedom
Degrees of freedom (df) is a concept in statistics that refers to the number of values in a calculation that are free to vary. It plays a critical role in various statistical tests, including determining the distribution of the test statistic.

In the context of our exercise, the degrees of freedom are derived using the formula \( df = n - 1 \), where \( n \) is the sample size. Here, with a sample size of 11 years, the degrees of freedom are 10. The degrees of freedom influence the shape of the chi-square distribution, impacting where the critical regions lie and ultimately the results of our hypothesis test.

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Most popular questions from this chapter

According to a public service website, \(69.4 \%\) of white collar criminals get prison time. A randomly selected sample of 165 white collar criminals revealed that 120 were serving or had served prison time. Using \(\alpha=0.05,\) test the conjecture that the proportion of white collar criminals serving prison time differs from \(69.4 \%\) in two different ways.

A researcher claims that the yearly consumption of soft drinks per person is 52 gallons. In a sample of 50 randomly selected people, the mean of the yearly consumption was 56.3 gallons. The standard deviation of the population is 3.5 gallons. Find the \(P\) -value for the test. On the basis of the \(P\) -value, is the researcher's claim valid?

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. Researchers suspect that \(18 \%\) of all high school students smoke at least one pack of cigarettes a day. At Wilson High School, a randomly selected sample of 300 students found that 50 students smoked at least one pack of cigarettes a day. At \(\alpha=0.05,\) test the claim that less than \(18 \%\) of all high school students smoke at least one pack of cigarettes a day. Use the \(P\) -value method.

A researcher wishes to see if the average number of sick days a worker takes per year is greater than \(5 .\) A random sample of 32 workers at a large department store had a mean of \(5.6 .\) The standard deviation of the population is \(1.2 .\) Is there enough evidence to support the researcher's claim at \(\alpha=0.01 ?\)

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. The amounts of vitamin C (in milligrams) for \(100 \mathrm{~g}\) ( 3.57 ounces) of various randomly selected fruits and vegetables are listed. Is there sufficient evidence to conclude that the standard deviation differs from \(12 \mathrm{mg}\) ? Use \(\alpha=0.10 .\) \(\begin{array}{rrrrrrr}7.9 & 16.3 & 12.8 & 13.0 & 32.2 & 28.1 & 34.4 \\ 46.4 & 53.0 & 15.4 & 18.2 & 25.0 & 5.2 & \end{array}\)
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