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Chapter 8: Problem 9

Researchers wish to test the efficacy of a program intended to reduce the length of labor in childbirth. The accepted mean labor time in the birth of a first child is 15.3 hours. The mean length of the labors of 13 first- time mothers in a pilot program was 8.8 hours with standard deviation 3.1 hours. Assuming a normal distribution of times of labor, test at the \(10 \%\) level of significance test whether the mean labor time for all women following this program is less than 15.3 hours.

Short Answer

Expert verified
The program significantly reduces mean labor time below 15.3 hours (\(p<0.10\)).

Step by step solution

01

State Hypotheses

First, we need to set up our null and alternative hypotheses. The null hypothesis (\(H_0\)) states that the mean labor time is 15.3 hours. The alternative hypothesis (\(H_a\)) suggests that the mean labor time is less than 15.3 hours: \[ H_0: \mu = 15.3 \] \[ H_a: \mu < 15.3 \] where \(\mu\) is the mean labor time for all women following this program.
02

Identify Given Information

From the problem, we know that the mean length of labor for the sample is 8.8 hours, the standard deviation is 3.1 hours, and the sample size is 13. The level of significance is \(\alpha = 0.10\).
03

Calculate the Test Statistic

Using the formula for the test statistic in a t-test: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \] Where \(\bar{x} = 8.8\) is the sample mean, \(\mu = 15.3\) is the population mean, \(s = 3.1\) is the sample standard deviation, and \(n = 13\) is the sample size. Substitute these values into the formula: \[ t = \frac{8.8 - 15.3}{3.1/\sqrt{13}} \] Calculating further gives \(t \approx -7.73\).
04

Determine Critical Value

Since the level of significance \(\alpha = 0.10\) and it's a one-tailed test, we look for the critical t-value with \(12\) degrees of freedom (\(n-1\)) in the t-distribution table. The critical value is approximately \(-1.356\).
05

Make Decision

Compare the calculated t-statistic with the critical value. Since \(-7.73 < -1.356\), we reject the null hypothesis \(H_0\).
06

Conclude

We have sufficient evidence at the 10% significance level to conclude that the mean labor time for women following this program is less than 15.3 hours.

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Key Concepts

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

t-test
The t-test is a statistical test used to compare the means of a sample to a known value, often the population mean. It's particularly useful when the sample size is small and the population standard deviation is unknown. In this exercise, the t-test helps determine if the average labor time for women in the program is significantly less than the usual 15.3 hours.
  • We use a t-test when dealing with normally distributed data and a relatively small sample size.
  • The formula for the t-test is: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \] where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.
The calculated test statistic tells us how far, in standard deviation units, the sample mean is from the population mean.
null hypothesis
The null hypothesis is a key concept in hypothesis testing. It represents the default position that there is no effect or no difference. In the context of this exercise, the null hypothesis (\(H_0\)) is that the mean labor time is equal to 15.3 hours.
  • The null hypothesis is what we aim to test against and potentially reject.
  • Formally stated, it is: \[ H_0: \mu = 15.3 \]
  • "Rejecting" the null hypothesis suggests that there is enough statistical evidence to support a different claim.
By setting a null hypothesis, we have a basis for comparison with the observed data.
alternative hypothesis
The alternative hypothesis is an assertion that contradicts the null hypothesis. It suggests the presence of an effect or difference. In our exercise, the alternative hypothesis (\(H_a\)) posits that the mean labor time is less than 15.3 hours.
  • The alternative hypothesis is what researchers hope to prove or find support for.
  • It is often considered as the opposite of the null, for instance: \[ H_a: \mu < 15.3 \]
  • We use hypothesis testing to determine whether the sample data provides sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
Successfully supporting the alternative hypothesis suggests a meaningful finding in the research.
critical value
The critical value is a threshold that the test statistic is compared to in hypothesis testing. It helps determine the statistical significance of the test result. For a one-sample t-test like in our exercise, the critical value is derived from the t-distribution table based on the level of significance and degrees of freedom.
  • In this context, with \( \alpha = 0.10 \) and 12 degrees of freedom, the critical value is approximately -1.356.
  • If the t-statistic falls beyond the critical value, the null hypothesis is rejected.
  • This critical comparison helps decide if we have enough evidence to support the alternative hypothesis.
In our exercise, the calculated t-statistic (-7.73) is more extreme than the critical value, leading to the rejection of the null hypothesis.

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Most popular questions from this chapter

Authors of a computer algebra system wish to compare the speed of a new computational algorithm to the currently implemented algorithm. They apply the new algorithm to 50 standard problems; it averages 8.16 seconds with standard deviation 0.17 second. The current algorithm averages 8.21 seconds on such problems. Test, at the \(1 \%\) level of significance, the alternative hypothesis that the new algorithm has a lower average time than the current algorithm.

Under what circumstance is a test of hypotheses certain to yield a correct decision?

State the null and alternative hypotheses for each of the following situations. (That is, identify the correct number \(\mu_{0}\) and write \(H_{0} \mu=\mu_{0}\) and the appropriate analogous expression for \(H_{a}\).) a. The average time workers spent commuting to work in Verona five years ago was 38.2 minutes. The Verona Chamber of Commerce asserts that the average is less now. b. The mean salary for all men in a certain profession is \(\$ 58,291\). A special interest group thinks that the mean salary for women in the same profession is different. c. The accepted figure for the caffeine content of an 8 -ounce cup of coffee is \(133 \mathrm{mg}\). A dietitian believes that the average for coffee served in a local restaurants is higher. d. The average yield per acre for all types of corn in a recent year was 161.9 bushels. An economist believes that the average yield per acre is different this year. e. An industry association asserts that the average age of all self-described fly fishermen is 42.8 years. A sociologist suspects that it is higher.

A medical laboratory claims that the mean turn-around time for performance of a battery of tests on blood samples is 1.88 business days. The manager of a large medical practice believes that the actual mean is larger. A random sample of 45 blood samples yielded mean 2.09 and sample standard deviation 0.13 day. Perform the relevant test at the \(10 \%\) level of significance, using these data.

The target temperature for a hot beverage the moment it is dispensed from a vending machine is \(170^{\circ} \mathrm{F}\). A sample of ten randomly selected servings from a new machine undergoing a pre- shipment inspection gave mean temperature \(173^{\circ} \mathrm{F}\) with sample standard deviation \(6.3^{\circ} \mathrm{F}\). a. Assuming that temperature is normally distributed, perform the test that the mean temperature of dispensed beverages is different from \(170^{\circ} \mathrm{F}\), at the \(10 \%\) level of significance. b. The sample mean is greater than 170 , suggesting that the actual population mean is greater than \(170^{\circ} \mathrm{F}\). Perform this test, also at the \(10 \%\) level of significance. (The computation of the test statistic done in part (a) still applies here.)
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