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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. High Temperatures in January Daily weather observations for southwestern Pennsylvania for the first three weeks of January for randomly selected years show daily high temperatures as follows: \(55,44,51,59,62,\) \(60,46,51,37,30,46,51,53,57,57,39,28,37,35,\) and 28 degrees Fahrenheit. The normal standard deviation in high temperatures for this time period is usually no more than 8 degrees. A meteorologist believes that with the unusual trend in temperatures the standard deviation is greater. At \(\alpha=0.05,\) can we conclude that the standard deviation is greater than 8 degrees?

Step by step solution

01

Define the Hypotheses

We want to test if the standard deviation of high temperatures is greater than 8 degrees. Our null hypothesis is \( H_0: \sigma = 8 \), where \( \sigma \) is the population standard deviation. The alternative hypothesis is \( H_a: \sigma > 8 \).

02

Gather the Data

The sample data of high temperatures is given as: 55, 44, 51, 59, 62, 60, 46, 51, 37, 30, 46, 51, 53, 57, 57, 39, 28, 37, 35, 28. Calculate the sample standard deviation from these values.

03

Calculate the Sample Variance and Standard Deviation

Calculate the sample mean and then use it to find the sample variance. The sample variance \( s^2 \) is calculated as: \[ s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2 \] where \( n \) is the sample size, \( x_i \) are the data points, and \( \bar{x} \) is the sample mean. Finally, take the square root to get the sample standard deviation \( s \).

04

Conduct the Hypothesis Test Using Chi-Square Distribution

Use the chi-square test for variance:\[ \chi^2 = \frac{(n-1)s^2}{\sigma^2} \] where \( n \) is the sample size and \( \sigma^2 \) is the population variance (64). Calculate \( \chi^2 \) using your variance from Step 3, with \( n = 20 \).

05

Determine the Critical Value

Since the hypothesis is right-tailed, use the chi-square distribution table to find the critical value for \( df = n-1 = 19 \) and \( \alpha = 0.05 \) which is approximately 30.144. This is the value where we will reject the null hypothesis if \( \chi^2 \) is greater.

06

Make the Decision

Compare the calculated \( \chi^2 \) with the critical value. If \( \chi^2 \) is greater than 30.144, reject the null hypothesis. Otherwise, do not reject the null hypothesis.

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Key Concepts

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

Standard Deviation

The term "standard deviation" is a measure that is used to quantify the amount of variation or dispersion in a set of data values. In simple terms, it is a statistic that tells you how spread out numbers are in your data set.
If the data points are close together, the standard deviation will be small. Conversely, if the data points are spread out over a wider range, the standard deviation will be larger.

• Standard deviation is calculated as the square root of the variance.
• It is denoted by the symbol \( s \) for a sample and \( \sigma \) for a population.

Understanding standard deviation aids in making predictions about data, helping to determine whether the spread of data is large or small. A small standard deviation indicates that the data points tend to be close to the mean, whereas a large standard deviation indicates that the data points are more spread out from the mean.

Chi-Square Distribution

When conducting hypothesis testing for variance, the chi-square distribution is especially valuable. It connects to the concept of variance in hypothesis testing and is fundamental in tests involving the square of normally distributed data.

• A chi-square distribution is not symmetric and skews to the right, especially for smaller degrees of freedom.
• It becomes more symmetric as the degrees of freedom increase.
• This distribution ranges from 0 to infinity.

For the given problem, when testing whether the standard deviation exceeds a certain threshold (like 8 degrees), the chi-square test uses the calculated chi-square value. This value is derived from the relationship: \[ \chi^2 = \frac{(n-1)s^2}{\sigma^2} \]Here \( n \) is the sample size, \( s^2 \) is the sample variance, and \( \sigma^2 \) is the known population variance.

Null Hypothesis

The null hypothesis, often denoted as \( H_0 \), is the statement being tested in a hypothesis test. It usually states that there is no effect or no difference, and it serves as the default or starting assumption.

• In our context, the null hypothesis is \( H_0: \sigma = 8 \), meaning the standard deviation of the high temperatures is equal to 8 degrees.
• The goal is to determine if there is enough evidence to reject the null hypothesis in favor of the alternative.
• Rejection of \( H_0 \) depends on whether the test statistic falls into the critical region defined by the distribution.

The null hypothesis essentially reflects the status quo or a baseline that one seeks to test against an alternative explanation.

Alternative Hypothesis

In hypothesis testing, the alternative hypothesis, denoted as \( H_a \), is what you want to prove or demonstrate. It's typically the statement you adopt if the evidence allows you to reject the null hypothesis.

• For the problem at hand, \( H_a: \sigma > 8 \) posits that the standard deviation of temperatures is greater than 8 degrees Fahrenheit.
• The alternative hypothesis challenges the null and is usually the statement reflecting a change or difference.
• It is what researchers and analysts are generally hoping to prove true through statistical evidence.

In hypothesis testing, confirming the alternative hypothesis involves showing that data significantly deviates from what is expected under the null hypothesis. Rejection of \( H_0 \) in favor of \( H_a \) would suggest a notable difference or effect.