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Answer the questions: (a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. A sample of 64 observations is selected from a normal population. The sample mean is \(215,\) and the population standard deviation is \(15 .\) Conduct the following test of hypothesis using the .025 significance level. $$ \begin{array}{l} H_{0}: \mu \geq 220 \\ H_{1}: \mu<220 \end{array} $$

Short Answer

Expert verified
(a) One-tailed. (b) Reject \(H_0\) if \(z < -1.96\). (c) \(z = -2.67\). (d) Reject \(H_0\). (e) \(p = 0.0038;\) evidence supports \(H_1\).

Step by step solution

01

Identify the Nature of the Test

The hypotheses provided are: - Null Hypothesis \(H_0: \mu \geq 220\)- Alternative Hypothesis \(H_1: \mu < 220\)The alternative hypothesis indicates that we are interested in checking if the mean is less than a certain value. Therefore, this is a one-tailed test, specifically a left-tailed test.
02

Set the Decision Rule

For a one-tailed test using a 0.025 significance level, we need to find the z-critical value where the cumulative probability is 0.025. Using z-tables or a calculator, the critical z-value is approximately -1.96. The decision rule is: - Reject \(H_0\) if the test statistic \(z < -1.96\).
03

Compute the Test Statistic

The test statistic for a hypothesis test concerning a population mean (with known standard deviation) is calculated as: \[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]Where:- \(\bar{x} = 215\) is the sample mean,- \(\mu = 220\) is the population mean under \(H_0\),- \(\sigma = 15\) is the standard deviation,- \(n = 64\) is the sample size.Calculate \(z\):\[z = \frac{215 - 220}{\frac{15}{\sqrt{64}}} = \frac{-5}{\frac{15}{8}} = \frac{-5}{1.875} = -2.67\]
04

Make a Decision Regarding H0

Compare the calculated test statistic (\(z = -2.67\)) with the critical value from the decision rule (\(-1.96\)). Since \(-2.67 < -1.96\), we reject the null hypothesis \(H_0\).
05

Calculate and Interpret the p-value

The p-value represents the probability of obtaining a test statistic as extreme as the observed, under the null hypothesis. For \(z = -2.67\), the cumulative probability from the z-table (or a calculator) is approximately 0.0038.Interpretation: The p-value of 0.0038 is less than the significance level of 0.025, which supports rejecting the null hypothesis \(H_0\).

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Key Concepts

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

One-tailed Test
In hypothesis testing, a one-tailed test is used when we are interested in determining if a parameter is significantly greater than or less than a certain value. In the given exercise, the alternative hypothesis tests if the mean is less than 220, which makes this a one-tailed test. Specifically, it is a left-tailed test because the direction is toward the lower tail of the distribution. This approach allows us to focus on one end of the distribution for more precise testing. A practical outcome of a one-tailed test is that it offers more power to detect an effect in one direction, as opposed to a two-tailed test, which considers deviations in both directions.

One-tailed tests are useful in instances where the outcome can only logically occur in one direction, or when such an outcome holds particular interest or significance to the researcher. By focusing on one direction, one-tailed tests enable researchers to make stronger inferences about the presence of an effect.
Critical Value
The critical value in hypothesis testing is the threshold against which the test statistic is compared to decide whether to reject the null hypothesis. For a one-tailed test at a 0.025 significance level, you find this value on a standard normal distribution table. In this case, the critical value for a left-tailed test is approximately -1.96.

The critical value is pivotal because it marks the point beyond which we consider the observed test statistic as unlikely under the null hypothesis. When the calculated test statistic falls beyond the critical value, we have sufficient evidence to reject the null hypothesis in favor of the alternative.

Understanding how to find and use the critical value is essential as it directly influences decision-making in statistical analysis by establishing a clear boundary for interpreting results.
Test Statistic
The test statistic is a standardized value used to compare against the critical value to determine the strength of the evidence against the null hypothesis. For means with a known population standard deviation, the test statistic is calculated using:

\[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]
  • \(\bar{x}\) is the sample mean.
  • \(\mu\) represents the hypothesized population mean under \(H_0\).
  • \(\sigma\) is the population standard deviation.
  • \(n\) is the sample size.

In the exercise, substituting in the values gives \(z = -2.67\). This test statistic tells us how many standard deviations the sample mean is from the hypothesized population mean.

The sign and magnitude of the test statistic are vital: the sign indicates the direction of the sample mean relative to the hypothesized mean, while the magnitude suggests how typical or atypical the sample is under the null hypothesis.
p-value
The p-value is a probability that measures the strength of the evidence against the null hypothesis. It represents the probability of obtaining a test statistic as extreme as, or more extreme than, the one calculated from the sample data, assuming the null hypothesis is true.

In this context, for a \(z\) test statistic of \(-2.67\), the p-value is approximately 0.0038. This is much smaller than the 0.025 significance level used in the test, indicating strong evidence against the null hypothesis.

Interpreting a p-value is straightforward:
  • A small p-value (typically \(<\) significance level) suggests rejecting \(H_0\).
  • A large p-value indicates insufficient evidence to reject \(H_0\).
Thus, a p-value provides a quantitative measure for deciding whether to reject the null hypothesis. The smaller the p-value, the stronger the evidence against \(H_0\).
Null Hypothesis
In hypothesis testing, the null hypothesis \((H_0)\) is a statement of no effect or no difference that serves as the starting point for statistical testing. It is the hypothesis that researchers typically aim to test against. In the exercise, \(H_0: \mu \geq 220\) suggests that the population mean is greater than or equal to 220.

The test aims to determine if there is enough statistical evidence to reject \(H_0\) in favor of the alternative hypothesis \((H_1)\), which in this case suggests \(\mu < 220\).

Null hypotheses are formulated to be straightforward and falsifiable. They provide a specific claim that can be tested with observed data, allowing researchers to use statistical methods to deduce if deviations from the claim are significant. It's important to note that rejecting \(H_0\) does not prove \(H_1\); it merely indicates that the data provide strong enough evidence against \(H_0\).

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Most popular questions from this chapter

The mean income per person in the United States is \(\$ 50,000,\) and the distribution of incomes follows a normal distribution. A random sample of 10 residents of Wilmington, Delaware, had a mean of \(\$ 60,000\) with a standard deviation of \(\$ 10,000 .\) At the .05 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average?

Rutter Nursery Company packages its pine bark mulch in 50 -pound bags. From a long history, the production department reports that the distribution of the bag weights follows the normal distribution, and the standard deviation of the packaging process is 3 pounds per bag. At the end of each day, Jeff Rutter, the production manager, weighs 10 bags and computes the mean weight of the sample. Below are the weights of 10 bags from today's production. \(\begin{array}{lllllllll}45.6 & 47.7 & 47.6 & 46.3 & 46.2 & 47.4 & 49.2 & 55.8 & 47.5 & 48.5\end{array}\) a. Can Mr. Rutter conclude that the mean weight of the bags is less than 50 pounds? Use the .01 significance level. b. In a brief report, tell why Mr. Rutter can use the \(z\) distribution as the test statistic. c. Compute the \(p\) -value.

A recent survey by nerdwallet.com indicated Americans paid a mean of \(\$ 6,658\) interest on credit card debt in 2017 . A sample of 12 households with children revealed the following amounts. At the .05 significance level, is it reasonable to conclude that these households paid more interest? \(\begin{array}{lllllllll}7,077 & 5,744 & 6,753 & 7,381 & 7,625 & 6,636 & 7,164 & 7,348 & 8,060 & 5,848 & 9,275 & 7,052\end{array}\)

The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.4 liters. A health campaign promotes the consumption of at least 2.0 liters per day. A sample of 10 adults after the campaign shows the following consumption in liters: $$\begin{array}{llllllll}1.5 & 1.6 & 1.5 & 1.4 & 1.9 & 1.4 & 1.3 & 1.9 & 1.8 & 1.7\end{array}$$ At the .01 significance level, can we conclude that water consumption has increased? Calculate and interpret the \(p\) -value.

Listed below is the annual rate of return (reported in percent) for a sample of 12 taxable mutual funds. \(\begin{array}{llllllllll}4.63 & 4.15 & 4.76 & 4.70 & 4.65 & 4.52 & 4.70 & 5.06 & 4.42 & 4.51 & 4.24 & 4.52\end{array}\) Using the .05 significance level, is it reasonable to conclude that the mean rate of return is more than \(4.50 \% ?\)

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