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Consider the null hypothesis \(H_{0}: \mu=12.80 .\) A random sample of 58 observations is taken from this population to perform this test. Using \(\alpha=.05\), show the rejection and nonrejection regions on the sampling distribution curve of the sample mean and find the critical value(s) of \(t\) for the following. a. a right-tailed test \(\underline{\text { b. a left-tailed test }}\) c. a two-tailed test

Short Answer

Expert verified
The critical values for a right-tailed test, left-tailed test and two-tailed test will be determined using a t-table for \(df=57\). The critical value for right-tailed test will be at \(\alpha=.05\) on the right of t-table, for left-tailed it'll be the same value but negative (at \(.05\) on the left of the table), while for the two-tailed test it will be at \(.025\) on both sides of the table.

Step by step solution

01

Understanding the concept

The null hypothesis presents a status quo or an assumption that there's no significant difference between the means of the population and the sample. The null hypothesis in this case is \(H_{0}: \mu=12.80\). Rejection and non-rejection regions are areas under the curve of a distribution; if the test statistic falls within the rejection region, the null hypothesis is rejected. The rejection region depends on the type of test (right-tailed, left-tailed, two-tailed) and the significance level, denoted by \(\alpha\). In this case, \(\alpha=.05\). The t-value is a test statistic that we compare with the critical value to determine whether to reject the null hypothesis.
02

Calculating for a right-tailed test

In a right-tailed test, the rejection region is to the right of the critical value, meaning we're testing if the mean is greater than \(H_{0}\). The degrees of freedom are calculated as the sample size minus one, \(df = 58 - 1 = 57\). Using a t-table, look for the value in the row of \(df=57\) and column \(\alpha=.05\). Whatever value you find is your critical value. The rejection area is any value larger than this critical t-value. The non-rejection region is to the left of this value.
03

Calculating for a left-tailed test

Contrarily to the right-tailed test, the rejection region for a left-tailed test is to the left of the critical value. We're testing if the mean is lesser than \(H_{0}\). Again, we do the same as in a right-tailed test but we now look at \(\alpha=.05\) for the left side. The critical value will be negative as the rejection region is on the left side of the mean. The non-rejection region is to the right of this value.
04

Calculating for a two-tailed test

Lastly, for a two tailed test, we're testing if the mean is either significantly lower or higher than \(H_{0}\). Therefore, the rejection region is both on the right and left of the distribution. For a two-tailed test, we divide the alpha value by 2, to distribute it equally on both sides, thus our new alpha is \(.05/2 = .025\). We look for t-value using \(df=57\) and \(\alpha=.025\) to find our positive critical value. The negative of this value is the critical value on the left. Anything beyond these values is the rejection region, and between them is the non-rejection region.

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Key Concepts

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

Understanding the t-distribution
The t-distribution is a type of probability distribution that is symmetrical and bell-shaped, similar to the normal distribution, but with heavier tails. This means it accounts for greater variability, especially in datasets with a smaller sample size. The tails of the t-distribution are thicker because it accommodates for the uncertainty in the sample variance, making it a perfect fit for situations where the sample size is less than 30 or the population standard deviation is unknown. The shape of the t-distribution depends on the degrees of freedom (df), which in most cases is the sample size minus one:
  • More degrees of freedom make the t-distribution closer to a normal distribution.
  • Fewer degrees of freedom make the t-distribution more spread out or flatter.
For hypothesis testing, particularly when assessing means, the t-distribution helps in determining the critical values, which aid in deciding whether to reject the null hypothesis.
Establishing the null hypothesis
The null hypothesis, denoted as \(H_{0}\), establishes a claim that there is no effect or no difference in the population; it represents the status quo. In the context of our exercise, the null hypothesis states that the population mean is 12.80. The purpose of setting up a null hypothesis is to provide a comparison framework for the test statistic. When we collect sample data, we calculate a test statistic to compare against critical values, derived from statistical tables like the t-table:
  • If the test statistic falls in the rejection region, we reject the null hypothesis.
  • If it falls in the non-rejection region, we do not reject it.
The null hypothesis thus provides the foundation upon which statistical tests are conducted, giving us a critical point of reference for our conclusions.
Determining the significance level
The significance level, often denoted as \(\alpha\), is a probability threshold set by the researcher before conducting the test. It represents the risk of rejecting the null hypothesis when it is actually true, commonly set at 0.05, 0.01, or 0.10.
  • In our exercise, \(\alpha = 0.05\) implies a 5% risk of making a Type I error, which is rejecting the null hypothesis when it should not be rejected.
  • The lower the \(\alpha\), the stricter we are, and the less likely we'll incorrectly reject \(H_{0}\).
This threshold is essentially used during hypothesis testing to determine the critical values that separate the rejection and non-rejection regions on the distribution curve. Selecting an appropriate significance level is crucial as it balances the risk of errors and the power we want our test to have.
Finding critical values
Critical values are the cutoff points on the test distribution that determine the start of the rejection region for the null hypothesis. They are determined by the significance level \(\alpha\) and the test being conducted (one-tailed or two-tailed).
  • In a right-tailed test, the critical value is found at \(\alpha\).
  • In a left-tailed test, the critical value is at \(-\alpha\).
  • In a two-tailed test, the critical values are found at \(\alpha/2\) and \(-\alpha/2\).
Using the t-distribution table and knowing the degrees of freedom, we can pinpoint these critical values, allowing us to scrutinize the null hypothesis. A sample mean falling beyond the critical values means rejecting \(H_{0}\), suggesting significant evidence against the initial claim of no difference or effect. Thus, critical values play an essential role in making objective decisions in hypothesis testing.

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Most popular questions from this chapter

A study claims that all adults spend an average of 14 hours or more on chores during a weekend. A researcher wanted to check if this claim is true. A random sample of 200 adults taken by this researcher showed that these adults spend an average of \(14.65\) hours on chores during a weekend. The population standard deviation is known to be \(3.0\) hours. a. Find the \(p\) -value for the hypothesis test with the alternative hypothesis that all adults spend more than 14 hours on chores during a weekend. Will you reject the null hypothesis at \(\alpha=.01 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01\).

A soft-drink manufacturer claims that its 12 -ounce cans do not contain, on average, more than 30 calories. A random sample of 64 cans of this soft drink, which were checked for calories, contained a mean of 32 calories with a standard deviation of 3 calories. Does the sample information support the altemative hypothesis that the manufacturer's claim is false? Use a significance level of \(5 \%\). Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value and \(\alpha=.05 ?\)

The manager of a service station claims that the mean amount spent on gas by its customers is $$\$ 15.90$$ per visit. You want to test if the mean amount spent on gas at this station is different from $$\$ 15.90$$ per visit. Briefly explain how you would conduct this test when \(\sigma\) is not known. \(9.73\) A tool manufacturing company claims that its top-of-the-line machine that is used to manufacture bolts produces an average of 88 or more bolts per hour. A company that is interested in buying this machine wants to check this claim. Suppose you are asked to conduct this test. Briefly explain how you would do so when \(\sigma\) is not known.

Perform the following tests of hypothesis. a. \(H_{0}: \mu=285, \quad H_{1}: \mu<285\) \(n=55, \quad \bar{x}=267.80, \quad s=42.90, \quad \alpha=.05\) b. \(H_{0-\mu}=10.70, \quad H_{1}: \mu \neq 10.70, \quad n=47, \bar{x}=12.025, \quad s=4.90, \quad \alpha=.01\) c. \(H_{0}=\mu=147,500, \quad H_{1}: \mu>147,500, n=41, \bar{x}=149,812, s=22,972, \alpha=.10\)

A paint manufacturing company claims that the mean drying time for its paints is not longer than 45 minutes. A random sample of 20 gallons of paints selected from the production line of this company showed that the mean drying time for this sample is \(49.50\) minutes with a standard deviation of 3 minutes. Assume that the drying times for these paints have a normal distribution. a. Using a \(1 \%\) significance level, would you conclude that the company's claim is true? b. What is the Type I error in this exercise? Explain in words. What is the probability of making such an error?

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