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Chapter 9: Problem 36

A random sample of 14 observations taken from a population that is normally distributed produced a sample mean of \(212.37\) and a standard deviation of \(16.35 .\) Find the critical and observed values of \(t\) and the range for the \(p\) -value for each of the following tests of hypotheses, using \(\alpha=.10\). a. \(H_{0}: \mu=205\) versus \(H_{1}: \mu \neq 205\) b. \(H_{0}: \mu=205\) versus \(H_{1}: \mu>205\)

Short Answer

Expert verified
The solution involves calculation of observed and critical t-values and determining the range for p-value. The null hypothesis can be rejected based on comparison of observed and critical t-values as well as p-value with alpha.

Step by step solution

01

Calculate t-value

The observed t鈥恦alue (t) is calculated using the following formula: \[ t = \frac{\bar{X} - \mu_0}{s / \sqrt{n}} \] Where: \(\bar{X}\) = sample mean = 212.37, \(\mu_0\) = population mean under \(H_0\) = 205 , s = sample standard deviation = 16.35 and n = sample size = 14.
02

Calculate critical t-value

The critical t-value distinguishes the acceptance region from rejection region. It is found from t-distribution table using degrees of freedom (n-1) and significance level \(\alpha\). For a two tail test, alpha/2 is used because the rejection regions are on both sides, while for a one tail test, alpha is used because rejection region is only on one side.
03

Range for P-value

P-value is the smallest level of significance at which null hypothesis could be rejected. P-value is found in conjunction with observed t-value. For two-tailed test, p-value equals the two tail areas beyond the observed t-value; for one-tailed test, p-value equals the one tail area beyond the observed t-value.
04

Interpret results

If the absolute value of t is greater than or equal to critical t-value, we reject the null hypothesis. Moreover, if p-value is less than alpha, we reject the null hypothesis.

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Key Concepts

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

Critical T-Value
A critical t-value helps us determine the threshold at which our findings become statistically significant. This value comes from the t-distribution table, which you can find using two key pieces of information: the degrees of freedom (df) and the significance level, \(\alpha\).
For our exercise, the degrees of freedom is calculated as the sample size minus one (n-1). In this example, with a sample size of 14, our degrees of freedom is 13.
  • Two-tailed test: For a significance level \(\alpha = 0.10\), we split the \(\alpha\) into 0.05 for each tail.
  • One-tailed test: Here, we maintain \(\alpha = 0.10\) for one direction of testing.
This critical t-value tells us where our rejection region lies. If our observed t-value goes beyond this critical value, we reject the null hypothesis.
P-Value
The p-value is a measure that helps us determine the strength of evidence against the null hypothesis. It represents the probability of observing the test results under the null hypothesis.
To understand it better:
  • If the p-value is small (less than \(\alpha\)), it suggests that the observed effect is strongly inconsistent with the null hypothesis.
  • If the p-value is large (greater than \(\alpha\)), it indicates that the evidence is not strong enough to reject the null hypothesis.
In our context, the p-value depends on whether we are conducting a one-tailed or a two-tailed test. For the two-tailed test, the p-value is the sum of both tail probabilities beyond the observed t-value, and for the one-tailed test, it accounts for only one side beyond this value.
Null Hypothesis
The null hypothesis is a statement we assume to be true before testing, and it is what we test against. It's a way of asserting that there's no effect or difference, often denoted by \(H_0\).
  • In our scenario, \(H_0: \mu = 205\) postulates that the mean of the population is 205.
  • The alternative hypothesis \(\H_1\) challenges this claim, suggesting a different population mean.
The test works to support or reject \(H_0\) based on the sample data. If the evidence points toward the improbability of \(H_0\), it leads us to reject it in favor of \(H_1\). This decision is informed by both the critical t-value and the p-value.
Sample Mean
The sample mean is a crucial statistic used in hypothesis testing. It represents the average value of a sample, noted as \(\bar{X}\).
  • In our example, the sample mean is given as 212.37, calculated from the 14 observations in the sample.
  • This value serves as a point estimate for the true population mean, \(\mu\).
Using the sample mean, alongside the sample standard deviation and the proposed mean under the null hypothesis, we calculate the observed t-value. The observed t-value helps indicate how far the sample mean is from the hypothetical population mean in units of standard error.

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Most popular questions from this chapter

A past study claimed that adults in America spent an average of 18 hours a week on leisure activities. A researcher wanted to test this claim. She took a sample of 12 adults and asked them about the time they spend per week on leisure activities. Their responses (in hours) are as follows. \(\begin{array}{lllllllllllll}13.6 & 14.0 & 24.5 & 24.6 & 22.9 & 37.7 & 14.6 & 14.5 & 21.5 & 21.0 & 17.8 & 21.4\end{array}\) Assume that the times spent on leisure activities by all American adults are normally distributed. Using a \(10 \%\) significance level, can you conclude that the average amount of time spent by American adults on leisure activities has changed? (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections 3.1.1 and \(3.2 .2\) of Chapter \(3 .\) Then make the test of hypothesis about \(\mu .\) )

An earlier study claimed that U.S. adults spent an average of 114 minutes per day with their family. A recently taken sample of 25 adults from a city showed that they spend an average of 109 minutes per day with their family. The sample standard deviation is 11 minutes. Assume that the times spent by adults with their families have an approximate normal distribution. a. Using a \(1 \%\) significance level, test whether the mean time spent currently by all adults with their families in this city is different from 114 minutes a day. b. Suppose the probability of making a Type I error is zero. Can you make a decision for the test of part a without going through the five steps of hypothesis testing? If yes, what is your decision? Explain.

Consider the null hypothesis \(H_{0}=p=.25 .\) Suppose a random sample of 400 observations is taken to perform this test about the population proportion. Using \(\alpha=.01\), show the rejection and nonrejection regions and find the critical value(s) of \(z\) for a a. left-tailed test b. two-tailed test c. right-tailed test

A study claims that \(65 \%\) of students at all colleges and universities hold off-campus (part-time or full-time) jobs. You want to check if the percentage of students at your school who hold off-campus jobs is different from \(65 \%\). Briefly explain how you would conduct such a test. Collect data from 40 students at your school on whether or not they hold off-campus jobs. Then, calculate the proportion of students in this sample who hold off-campus jobs. Using this information, test the hypothesis. Select your own significance level.

For each of the following examples of tests of hypotheses about \(\mu\), show the rejection and nonrejection regions on the sampling distribution of the sample mean assuming that it is normal. a. A two-tailed test with \(\alpha=.05\) and \(n=40\) b. A left-tailed test with \(\alpha=.01\) and \(n=20\) c. A right-tailed test with \(\alpha=.02\) and \(n=55\)
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