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Chapter 8: Problem 11

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Tornado Deaths 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; there isn't enough evidence to support that the standard deviation is less than 35.

Step by step solution

01

State the Hypotheses

We need to determine the null and alternative hypotheses for this test. The null hypothesis, denoted as \(H_0\), assumes that the standard deviation, \(\sigma\), is equal to \(35\), while the alternative hypothesis, denoted as \(H_1\), suggests that \(\sigma\) is less than \(35\). We write this as:\[ H_0: \sigma = 35 \]\[ H_1: \sigma < 35 \]
02

Determine the Test Statistic

To test the hypothesis about the standard deviation, we use the chi-square distribution. The test statistic is given by:\[ \chi^2 = \frac{(n-1)s^2}{\sigma_0^2} \]where \(n = 11\) is the sample size, \(s = 32\) is the sample standard deviation, and \(\sigma_0 = 35\) is the hypothesized standard deviation under the null hypothesis. Plugging in the values:\[ \chi^2 = \frac{(11-1) \times 32^2}{35^2} \]
03

Calculate the Chi-Square Statistic

Evaluate the expression for \(\chi^2\):\[ \chi^2 = \frac{10 \times 1024}{1225} \approx 8.34 \]
04

Determine the Critical Value

To find the critical value, we need the chi-square distribution table for \(n-1 = 10\) degrees of freedom, considering \(\alpha = 0.05\) for a left-tailed test. The critical value for \(\chi^2\) with 10 degrees of freedom at \(\alpha = 0.05\) is approximately \(5.89\).
05

Make the Decision

Compare the calculated \(\chi^2\) value to the critical value. Since \(8.34\) is greater than \(5.89\), we do not reject the null hypothesis.
06

Conclusion

Since we did not reject \(H_0\), there is not enough statistical evidence to support the claim that the standard deviation of the number of deaths annually from tornadoes in the United States 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 crucial part of hypothesis testing, especially when analyzing variations or the standard deviations of a dataset. It's used when you need to test hypotheses regarding the variability of a population. In this context, the chi-square distribution helps determine whether the observed variance differs significantly from the hypothetical variance. This distribution is skewed to the right and is defined by the degrees of freedom, which is usually determined by the sample size minus one (n-1).
  • In hypothesis tests, specifically those about standard deviation, we use the chi-square statistic formula: \( \chi^2 = \frac{(n-1)s^2}{\sigma_0^2} \) where \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\sigma_0\) is the population standard deviation under the null hypothesis.
  • Chi-square values are compared to a critical value from a chi-square distribution table, specific to the confidence level and degrees of freedom.
  • If the calculated chi-square statistic is greater than the critical value in a left-tailed test, we don't reject the null hypothesis, as seen in our exercise.
Understanding and correctly applying the chi-square distribution ensures accurate interpretations and decisions in statistical analyses.
Statistical Evidence
When we refer to statistical evidence in hypothesis testing, we're talking about the probability that the observed data occurred under the null hypothesis. It helps us determine whether to accept or reject the null hypothesis.
  • Statistical evidence is gathered by calculating a test statistic - in this case, a chi-square statistic - and comparing it to a critical value from a statistical table.
  • The evidence is quantified in terms of the p-value, which measures the strength of the evidence against the null hypothesis. However, in this exercise, we primarily rely on critical values from chi-square distributions for the decision.
Essentially, strong statistical evidence against the null hypothesis occurs when the calculated test statistic significantly deviates under given experimental conditions.
In our exercise, the computed chi-square statistic did not provide enough evidence to support the researcher's claim, emphasizing how evidence-based decisions are drawn in statistical testing.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It's a key player in determining how data points spread out from the mean, and it's crucial when conducting hypothesis tests about data variability.
  • A low standard deviation indicates that data points tend to be close to the mean, while a high standard deviation indicates more spread out data.
  • In our scenario, the researcher's claim is centered around whether the standard deviation of tornado-related deaths is less than 35. Calculating the sample standard deviation (32 in this case) gives us an initial idea of the dataset's variability.
  • Our hypothesis test aims to see if the sample standard deviation significantly deviates from the hypothesized value under the null hypothesis.
Understanding standard deviation is essential, as it's not just a measure for variation but also a key input in various statistical formulas and distribution calculations, forming the basis of evidenced conclusions in research.
Null Hypothesis
The null hypothesis is a foundational concept in statistics, representing a statement or default position that there is no effect or no difference, until proven otherwise. It provides a starting point for hypothesis testing.
  • In our exercise, the null hypothesis \( (H_0) \) suggests that the standard deviation of tornado deaths is equal to 35.
  • This hypothesis is challenged by the alternative hypothesis \( (H_1) \) that claims the standard deviation is actually less than 35.
  • Hypothesis testing involves using statistical methods to either reject or fail to reject \(H_0\), based on data evidence.
The null hypothesis is central because it encapsulates the status quo, which must be disproven to substantiate the researcher's claim.
In conclusion, while the step-by-step solution in our exercise pointed out that the null hypothesis couldn't be rejected, it acts as a steady benchmark against which alternative claims are measured.

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Most popular questions from this chapter

Working at Home Workers with a formal arrangement with their employer to be paid for time worked at home worked an average of 19 hours per week. A random sample of 15 mortgage brokers indicated that they worked a mean of 21.3 hours per week at home with a standard deviation of 6.5 hours. At $$\alpha=0.05,$$ is there sufficient evidence to conclude a difference? Construct a $$95 \%$$ confidence interval for the true mean number of paid working hours at home. Compare the results of your confidence interval to the conclusion of your hypothesis test and discuss the implications.

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Calories in Pancake Syrup A nutritionist claims that the standard deviation of the number of calories in 1 tablespoon of the major brands of pancake syrup is \(60 .\) A random sample of major brands of syrup is selected, and the number of calories is shown. At \(\alpha=0.10,\) can the claim be rejected? $$ \begin{array}{cccccc}{53} & {210} & {100} & {200} & {100} & {220} \\ {210} & {100} & {240} & {200} & {100} & {210} \\ {100} & {210} & {100} & {210} & {100} & {60}\end{array} $$

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Transferring Phone Calls The manager of a large company claims that the standard deviation of the time (in minutes) that it takes a telephone call to be transferred to the correct office in her company is 1.2 minutes or less. A random sample of 15 calls is selected, and the calls are timed. The standard deviation of the sample is 1.8 minutes. At \(\alpha=0.01\), test the claim that the standard deviation is less than or equal to 1.2 minutes. Use the \(P\) -value method.

Using Table \(F,\) find the \(P\) -value interval for each test value. a. \(t=3.025, n=24,\) right-tailed b. \(t=-1.145, n=5,\) left-tailed c. \(t=2.179, n=13,\) two-tailed d. \(t=0.665, n=10,\) right-tailed

For Exercises I through 25, 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 diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Farm Sizes Ten years ago, the average acreage of farms in a certain geographic region was 65 acres. The standard deviation of the population was 7 acres. A recent study consisting of 22 randomly selected farms showed that the average was 63.2 acres per farm. Test the claim, at \(\alpha=0.10,\) that the average has not changed by finding the \(P\) -value for the test. Assume that \(\sigma\) has not changed and the variable is normally distributed.
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