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Chapter 8: Problem 1

First-Time Births According to the almanac, the mean age for a woman giving birth for the first time is 25.2 years. A random sample of ages of 35 professional women giving birth for the first time had a mean of 28.7 years and a standard deviation of 4.6 years. Use both a confidence interval and a hypothesis test at the 0.05 level of significance to test if the mean age of professional woman is different from 25.2 years at the time of their first birth.

Short Answer

Expert verified
Yes, the mean age is significantly different from 25.2 years.

Step by step solution

01

Define Null and Alternative Hypotheses

To test if the mean age of professional women giving birth for the first time is different from 25.2 years, we set up our hypotheses. \( H_0 \): The mean age \( \mu = 25.2 \) years. \( H_a \): The mean age \( \mu eq 25.2 \) years. This is a two-tailed test.
02

Calculate Test Statistic

The test statistic (t) is calculated using the formula: \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \), where \( \bar{x} = 28.7 \), \( \mu = 25.2 \), \( s = 4.6 \), and \( n = 35 \). Calculating, \( t = \frac{28.7 - 25.2}{4.6/\sqrt{35}} \approx 4.705 \).
03

Determine Critical Value and Analyze Results

With degrees of freedom \( df = n - 1 = 34 \), and a significance level of 0.05, we find the critical t-value corresponding to a two-tailed test. Looking up the t-distribution table or using a calculator gives \( t_{critical} \approx 2.032 \). Since \( |t| = 4.705 > 2.032 \), we reject \( H_0 \).
04

Construct the Confidence Interval

To construct a 95% confidence interval for the mean, use the formula: \( \bar{x} \pm t_{critical} \times \frac{s}{\sqrt{n}} \). Substituting the values, the confidence interval is \( 28.7 \pm 2.032 \times \frac{4.6}{\sqrt{35}} \approx 28.7 \pm 1.58 \), giving the interval (27.12, 30.28).
05

Interpret the Results

The confidence interval (27.12, 30.28) does not include 25.2, supporting our test result. Both the hypothesis test and the confidence interval suggest that the mean age of professional women giving birth for the first time is significantly different from 25.2 years.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about a population based on a sample. In this scenario, we want to test if the mean age at which professional women give birth for the first time is different from the general population, described here as 25.2 years. To do this, we set up two hypotheses:
  • The null hypothesis (\( H_0 \)): The mean age \( \mu = 25.2 \) years, implying no difference.
  • The alternative hypothesis (\( H_a \)): The mean age \( \mu eq 25.2 \) years, suggesting a difference.
We perform a two-tailed test because we're looking for any difference, be it higher or lower than 25.2 years. A sample mean of 28.7 years indicates a potential difference, which we evaluate statistically using a t-test. If the test statistic exceeds the critical value, we reject \( H_0 \), concluding the sample mean is significantly different.
Confidence Interval
A confidence interval provides a range within which we expect the true population parameter to lie, with a certain level of confidence. In our example, we calculated a 95% confidence interval to understand the range of the mean age for first-time births among professional women.
  • The formula used is: \( \bar{x} \pm t_{critical} \times \frac{s}{\sqrt{n}} \).
  • Our sample mean (\( \bar{x} \)) is 28.7, with a standard deviation (\( s \)) of 4.6 and sample size (\( n \)) of 35.
  • With the critical t-value at 2.032, the confidence interval is approximately (27.12, 30.28).
This interval does not contain the hypothesized population mean of 25.2, supporting the result of our hypothesis test, and indicating a significant difference in the mean age for first-time births among professional women.
Mean Age
The mean age, or average age, is a key part of this analysis. It represents the central point of our data. For our sample of professional women, the mean age at which they first give birth is calculated to be 28.7 years, which is noticeably higher than the population mean of 25.2 years. By comparing this sample mean with the hypothetical population mean, we can infer whether there's a significant difference.
  • The mean provides a simple measure of central tendency.
  • It allows us to make comparisons between different groups or populations.
In the context of hypothesis testing, the sample mean is crucial in determining whether the deviation from the expected mean (25.2 years) is statistically significant.
Significance Level
The significance level in hypothesis testing determines how unlikely a result must be, under the null hypothesis, before we consider it evidence against the null. Conventional levels are set at 0.05, balancing between Type I errors (false positives) and practical significance.
  • In our example, a 0.05 significance level is used, equating to a 5% risk of incorrectly rejecting the null hypothesis.
  • This level defines the critical region 鈥� values of the test statistic falling in this region lead to rejection of \( H_0 \).
  • Our computed test statistic for the mean age is 4.705, well beyond the critical value of 2.032 for the t-distribution with 34 degrees of freedom.
Thus, at a 0.05 significance level, the observed mean age provides strong evidence against the null hypothesis, confirming a genuine difference in age for first-time births among professional women compared to the general population.

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Most popular questions from this chapter

For Exercises 7 through \(23,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. Commute Time to Work A survey of 15 large U.S. cities finds that the average commute time one way is 25.4 minutes. A chamber of commerce executive feels that the commute in his city is less and wants to publicize this. He randomly selects 25 commuters and finds the average is 22.1 minutes with a standard deviation of 5.3 minutes. At \(\alpha=0.10,\) is he correct?

In hypothesis testing, why can't the hypothesis be proved true?

Significance Levels Suppose a statistician chose to test a hypothesis at \(\alpha=0.01 .\) The critical value for a right-tailed test is \(+2.33 .\) If the test value were 1.97 , what would the decision be? What would happen if, after seeing the test value, she decided to choose \(\alpha=0.05 ?\) What would the decision be? Explain the contradiction, if there is one.

For Exercises 7 through \(23,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. Television Viewing by Teens Teens are reported to watch the fewest total hours of television per week of all the demographic groups. The average television viewing for teens on Sunday from 1: 00 to 7: 00 P.M. is 58 minutes. A random sample of local teens disclosed the following times for Sunday afternoon television viewing. At \(\alpha=0.01,\) can it be concluded that the average is greater than the national viewing time? (Note: Change all times to minutes.) $$ \begin{array}{llll}{2: 30} & {2: 00} & {1: 30} & {3: 20} \\ {1: 00} & {2: 15} & {1: 50} & {2: 10} \\ {1: 30} & {2: 30} & {}\end{array} $$

Newspaper Reading Times A survey taken several years ago found that the average time a person spent reading the local daily newspaper was 10.8 minutes. The standard deviation of the population was 3 minutes. To see whether the average time had changed since the newspaper's format was revised, the newspaper editor surveyed 36 individuals. The average time that these 36 randomly selected people spent reading the paper was 12.2 minutes. At $$\alpha=0.02,$$ is there a change in the average time an individual spends reading the newspaper? Find the $$98 \%$$ confidence interval of the mean. Do the results agree? Explain.
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