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Chapter 8: Problem 10

A random sample has 49 values. The sample mean is 8.5 and the sample standard deviation is \(1.5 .\) Use a level of significance of 0.01 to conduct a left- tailed test of the claim that the population mean is \(9.2 .\) (a) Check Requirements Is it appropriate to use a Student's \(t\) distribution? Explain. How many degrees of freedom do we use? (b) What are the hypotheses? (c) Compute the \(t\) value of the sample test statistic. (d) Estimate the \(P\) -value for the test. (e) Do we reject or fail to reject \(H_{0} ?\) (f) Interpret the results.

Short Answer

Expert verified
We reject \( H_0 \) and conclude \( \mu < 9.2 \).

Step by step solution

01

Check Requirements

To determine if it's appropriate to use a Student's \( t \) distribution, we need to check if the sample size is sufficiently large or if the population is normally distributed. The sample size is 49, which is larger than 30, making it appropriate to use the \( t \) distribution. The degrees of freedom is calculated as \( n - 1 = 49 - 1 = 48 \).
02

State the Hypotheses

We perform a left-tailed test. The null hypothesis \( H_0 \) is that the population mean \( \mu = 9.2 \). The alternative hypothesis \( H_a \) is that the population mean \( \mu < 9.2 \). So, \( H_0: \mu = 9.2 \) and \( H_a: \mu < 9.2 \).
03

Calculate the Test Statistic

The test statistic for a \( t \) test is calculated using the formula: \[ t = \frac{\bar{x} - \mu}{(s/\sqrt{n})} \] where \( \bar{x} = 8.5 \), \( \mu = 9.2 \), \( s = 1.5 \), and \( n = 49 \). Plugging in these values gives: \[ t = \frac{8.5 - 9.2}{1.5/\sqrt{49}} = \frac{-0.7}{1.5/7} = \frac{-0.7}{0.2143} \approx -3.26 \]
04

Estimate the P-value

Using the \( t \) distribution table (or software) for \( t = -3.26 \) and \( 48 \) degrees of freedom, we estimate the \( P \)-value. The \( P \)-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed value under the null hypothesis. Given \( t = -3.26 \), the \( P \)-value is less than 0.01 (specifically, it's close to 0.001).
05

Decision Rule

Since the \( P \)-value (\(< 0.01\)) is less than the significance level \( \alpha = 0.01 \), we reject the null hypothesis \( H_0 \).
06

Interpret the Results

By rejecting \( H_0 \), we conclude that there is sufficient evidence to support the claim that the population mean \( \mu \) is less than 9.2 at the 0.01 significance level.

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Key Concepts

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

Student's t-distribution
In statistics, the Student's t-distribution is a fundamental concept used to carry out hypothesis testing when the sample size is small and/or the population standard deviation is unknown. It is particularly useful because it allows us to make inferences about a population mean in situations where the normal distribution would not be appropriate.

Key features of the t-distribution include:
  • The t-distribution is used instead of the normal distribution primarily when dealing with small sample sizes (typically less than 30) or when the population variance is unknown.
  • Its shape is similar to the normal distribution but with heavier tails. This means it accounts for the greater variability expected with smaller samples.
  • As the sample size increases, the t-distribution approaches the normal distribution (Central Limit Theorem).
  • The specific form of the distribution is determined by the 'degrees of freedom' which will be discussed later.
In our exercise, with a sample size of 49, it's acceptable to use the t-distribution because even though it's not small, the population standard deviation is unknown.
Left-tailed test
A left-tailed test is a type of hypothesis test where the area of interest is on the left side of the distribution. This means we are testing the likelihood that the population mean is less than a certain value.

In the context of hypothesis testing:
  • The null hypothesis (\( H_0 \)) states that there's no effect or difference. For our exercise, it claims the population mean is equal to 9.2.
  • The alternative hypothesis (\( H_a \)) suggests that there is a departure from the null hypothesis, in this case, that the population mean is less than 9.2.
  • "Left-tailed" refers to the alternative hypothesis suggesting a mean less than the claimed value. Our sample mean being considerably lower (8.5 vs 9.2) justifies this test.
The critical region, or rejection area, is on the left part of the t-distribution, leading to rejection of the null hypothesis if the test statistic is sufficiently small or negative.
Degrees of freedom
Degrees of freedom are a concept in statistics that describe the number of values in a calculation that are free to vary. When conducting a t-test, the degrees of freedom refer to the number of independent data points that went into the calculation of the sample variance.

For our sample:
  • The formula to calculate degrees of freedom is usually \( n - 1 \), where \( n \) is the sample size. This subtracted 1 accounts for the estimation of the sample mean from the data.
  • In our exercise, with a sample of 49, the degrees of freedom are calculated as \( 49 - 1 = 48 \).
  • This value is crucial as it determines the shape of the Student's t-distribution used for the hypothesis test.
  • In general, more degrees of freedom (i.e., a larger sample size) lead to a t-distribution that closely resembles a normal distribution.
P-value estimation
The P-value is a fundamental concept in hypothesis testing that helps us decide whether to reject the null hypothesis. It reflects the probability of obtaining a test statistic as extreme as the one observed, assuming the null hypothesis is true.

In our scenario:
  • A low P-value (typically 鈮� 0.05) indicates strong evidence against the null hypothesis, thus leading us to reject it. For our left-tailed test, the calculated t-statistic was approximately \( -3.26 \).
  • Utilizing the degrees of freedom (48) and the test statistic, we can either use a t-distribution table or statistical software to estimate the P-value.
  • In this case, the P-value is noted to be less than 0.01, which is the significance level set for our test, showing clear evidence against \( H_0 \).
Deciding to reject or fail to reject \( H_0 \) hinges on this P-value. A P-value less than the significance level of 0.01 means we reject the null hypothesis, suggesting that the population mean is truly less than 9.2.

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Most popular questions from this chapter

Please provide the following information for Problems. (a) What is the level of significance? State the null and alternate hypotheses. (b) Check Requirements What sampling distribution will you use? What assumptions are you making? Compute the sample test statistic and corresponding \(z\) or \(t\) value as appropriate. (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha ?\) (e) Interpret your conclusion in the context of the application. Note: For degrees of freedom \(d . f .\) not in the Student's \(t\) table, use the closest \(d . f .\) that is smaller. In some situations, this choice of \(d . f .\) may increase the \(P\) -value a small amount and therefore produce a slightly more "conservative" answer. Medical: Hay Fever A random sample of \(n_{1}=16\) communities in western Kansas gave the following information for people under 25 years of age. \(x_{1}:\) Rate of hay fever per 1000 population for people under 25 \(\begin{array}{cccccccc}98 & 90 & 120 & 128 & 92 & 123 & 112 & 93 \\ 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88\end{array}\) A random sample of \(n_{2}=14\) regions in western Kansas gave the following information for people over 50 years old. \(x_{2}:\) Rate of hay fever per 1000 population for people over 50 \(\begin{array}{ccccccc}95 & 110 & 101 & 97 & 112 & 88 & 110\end{array}\) \(\begin{array}{ccccccc}79 & 115 & 100 & 89 & 114 & 85 & 96\end{array}\) (Reference: National Center for Health Statistics.) i. Use a calculator to verify that \(\bar{x}_{1} \approx 109.50, s_{1} \approx 15.41, \bar{x}_{2} \approx 99.36,\) and \(s_{2} \approx 11.57\) ii. Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use \(\alpha=0.05\)

Paired Differences Test For a random sample of 36 data pairs, the sample mean of the differences was 0.8. The sample standard deviation of the differences was \(2 .\) At the \(5 \%\) level of significance, test the claim that the population mean of the differences is different from 0. (a) Check Requirements Is it appropriate to use a Student's \(t\) distribution for the sample test statistic? Explain. What degrees of freedom are used? (b) State the hypotheses. (c) Compute the sample test statistic and corresponding \(t\) value. (d) Estimate the \(P\) -value of the sample test statistic. (e) Do we reject or fail to reject the null hypothesis? Explain. (f) Interpretation What do your results tell you?

Consider a hypothesis test of difference of proportions for two independent populations. Suppose random samples produce \(r_{1}\) successes out of \(n_{1}\) trials for the first population and \(r_{2}\) successes out of \(n_{2}\) trials for the second population. What is the best pooled estimate \(\bar{p}\) for the population probability of success using \(H_{0}: p_{1}=p_{2} ?\)

(a) What is the level of significance? State the null and alternate hypotheses. (b) Check Requirements What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. Compute the appropriate sampling distribution value of the sample test statistic. (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha ?\) (e) Interpret your conclusion in the context of the application. Note: For degrees of freedom \(d . f .\) not given in the Student's \(t\) table, use the closest \(d . f .\) that is smaller. In some situations, this choice of \(d . f .\) may increase the \(P\) -value by a small amount and therefore produce a slightly more "conservative" answer. Let \(x\) be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the \(x\) distribution is \(\mu=7.4\) (Reference: The Merck Manual, a commonly used reference in medical schools and nursing programs). A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 31 patients with arthritis took the drug for 3 months. Blood tests showed that \(\bar{x}=8.1\) with sample standard deviation \(s=1.9 .\) Use a \(5 \%\) level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood.

Weatherwise magazine is published in association with the American Meteorological Society. Volume \(46,\) Number 6 has a rating system to classify Nor'easter storms that frequently hit New England states and can cause much damage near the ocean coast. A severe storm has an average peak wave height of 16.4 feet for waves hitting the shore. Suppose that a Nor'easter is in progress at the severe storm class rating. (a) Let us say that we want to set up a statistical test to see if the wave action (i.e., height) is dying down or getting worse. What would be the null hypothesis regarding average wave height? (b) If you wanted to test the hypothesis that the storm is getting worse, what would you use for the alternate hypothesis? (c) If you wanted to test the hypothesis that the waves are dying down, what would you use for the alternate hypothesis? (d) Suppose you do not know whether the storm is getting worse or dying out. You just want to test the hypothesis that the average wave height is different (either higher or lower) from the severe storm class rating. What would you use for the alternate hypothesis? (e) For each of the tests in parts (b), (c), and (d), would the area corresponding to the \(P\) -value be on the left, on the right, or on both sides of the mean? Explain your answer in each case.
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