/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 In a study on the fertility of m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a study on the fertility of married women conducted by Martin O'Connell and Carolyn C. Rogers for the Census Bureau in 1979 , two groups of childless wives aged 25 to 29 were selected at random and each wife was asked if she eventually planned to have a child. One group was selected from among those wives married less than two years and the other from among those wives married five years. Suppose that 240 of 300 wives married less than two years planned to have children some day compared to 288 of the 400 wives married five years, Can we conclude that the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years? Make use of a P-value

Short Answer

Expert verified
Based on the test, we conclude that the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years. Our p-value is 0.0062, which is less than the usual significant level of 0.05.

Step by step solution

01

Calculate the Proportions

First, calculate the proportion of wives from each group who plan to have children. To do this, divide the number of affirmative responses by the total number of wives in each group. For those married less than two years, the proportion is \( \frac{240}{300} = 0.8 \) and for those married five years, the proportion is \( \frac{288}{400} = 0.72 \).
02

Formulate Hypotheses

The next step is setting up the null and alternative hypotheses. Let's use \(p_{1}\) and \(p_{2}\) to represent the proportions of wives from each group. The null hypothesis \(H_{0}\) would be \(p_{1} = p_{2}\), meaning there is no difference between the proportions. The alternative hypothesis \(H_{1}\) would be \(p_{1} > p_{2}\), i.e., the proportion of wives married less than two years who plan to have children is greater than that of wives married five years.
03

Calculate Standard Error

Next, calculate the standard error (SE) for the difference in proportions. This can be calculated using the formula SE \(= \sqrt{\frac{p_{1}*(1-p_{1})}{n_{1}} + \frac{p_{2}*(1-p_{2})}{n_{2}}}\). With \(n_{1} = 300\), \(n_{2} = 400\), \(p_{1} = 0.8\), and \(p_{2} = 0.72\), the SE becomes approximately 0.032.
04

Compute test statistic

The test statistic z, is calculated by \(z = \frac{p_{1} - p_{2}}{SE}\). Substituting our values, we get \(z = \frac{0.8 - 0.72}{0.032} = 2.5\).
05

P-value Calculation and Conclusion

Finally, compare the calculated z-value (2.5) to the critical z-value (typically 1.645 for a one-tailed test with a significance level of 0.05). Since the calculated z-value is greater than the critical z-value, we reject the null hypothesis. The P-value in this case is 0.0062, which confirms our rejection of the null hypothesis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Proportions
Proportions give us insight into the part of a whole, especially useful in statistical testing. In the exercise, we calculate proportions to determine the fraction of wives in each group who plan to have children. For those married less than two years, we find it's 0.8, while for those married for five years, it's 0.72.
By comparing these proportions, we attempt to see if there's a significant difference in their intents.
Proportions are important because they allow us to compare different groups on a scale that is relatable and standardized, no matter the group size.
  • Proportions are calculated by dividing the number interested or successful outcomes by the total number of outcomes.
  • They serve as the basis for later analysis, especially when comparing different population segments.
  • Allow us to set up hypotheses to further statistical investigation.
Standard Error
The Standard Error (SE) is crucial for understanding the variability of a statistic from sample data. It essentially measures how much sample statistics, like proportions, deviate from their true population values.
In simpler terms, SE tells us how "spread out" the sample data is likely to be from the actual data.
In hypothesis testing, SE helps determine the reliability of the difference between two sample statistics.

Standard Error Calculation

For the problem, SE is used to find the variability in the difference between two proportions. It helps assess the degree to which sample proportions might differ by chance alone.
The formula used here is:
\[ SE = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}} \]
Using this method gives us a value of approximately 0.032 in the solution, indicating the variability around the difference of proportions.
  • Helps to interpret how "statistically significant" the results are.
  • A smaller SE suggests the sample statistic is a more precise estimate of the population parameter.
P-value
The P-value is a central concept in hypothesis testing. It helps us decide whether to accept or reject a hypothesis by providing a measure of evidence against the null hypothesis.
It represents the probability of observing our results—or something more extreme—assuming that the null hypothesis is true.

Significance of P-value

In our exercise, a calculated P-value of 0.0062 indicates that such a large or larger difference in proportions would be very rare if the null hypothesis were correct.
Thus, a small P-value (typically less than 0.05) leads us to reject the null hypothesis. Here, this supports the belief that there is indeed a significant difference in the childbearing plans of the two groups.
  • A P-value provides a method to conclude whether sample findings are unusual under the null hypothesis.
  • Lower P-values mean stronger evidence against the null hypothesis, suggesting a significant difference.
  • An essential tool in making data-driven decisions.
Z-test
The Z-test is used when we want to test the significance of the difference between two population proportions.
It works well with large sample sizes and known variances.

Using a Z-test in Hypotheses

In the exercise, the Z-test examines the hypothesis that recently married wives are more likely to plan for children than those married longer.
The Z-value is computed using:
\[ z = \frac{p_1 - p_2}{SE} \]
With our calculated Z-value of 2.5, the corresponding P-value supports rejecting the null hypothesis.
  • Empowers us to statistically validate whether one group's characteristics significantly differ from another's.
  • Useful with known standard deviations and larger sample sizes.
  • The Z-value converts a difference in proportions into a standardized metric we can compare to known statistical tables.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

State the null and alternative hypotheses to be used in testing the following claims and determine generally where the critical region is located: (a) The mean snowfall at Lake George during the month of February is 21.8 centimeters. (b) No more than \(20 \%\) of the faculty at the local university contributed to the annual giving fund. (c) On the average, children attend schools within 6.2 kilometers of their homes in suburban St. Louis. (d) At least \(70 \%\) of next year's new cars will be in the compact and subcompact category. (e) The proportion of voters favoring the incumbent in the upcoming election is 0.58 . (f) The average rib-eye steak at the I.onghorn Steak house is at least 340 grams.

10.52 On testing $$\begin{array}{l}\text { Ho: } p=14 \\\H_{1}: \quad \mu \neq 14\end{array}$$ an \(\alpha=0.05\) level \(t\) -test is being considered. What sample size is necessary in order that the probability is 0.1 of falsely failing to reject \(H o\) when the true population mean differs from 14 by \(0.5 ?\) From a preliminary sample we estimate \(\sigma\) to be \(1.25 .\)

A manufacturer has developed a new fishing line, which he claims has a mean breaking strength of 15 kilograms with a standard deviation of 0.5 kilogram. To test the hypothesis that \(\mu=15\) kilograms against the alternative that \(p<15\) kilograms, a random sample of 50 lines will be tested. The critical region is defined to be \(x<14.9\) (a) Find the probability of committing a type 1 error when \(H_{0}\) is true (b) Evaluate \(\beta\) for the alternatives \(p-14.8\) and \(\mu=\) 14.9 kilograms.

Al. a certain college it is estimated that at most \(25 \%\) of the students ride bicycles to class. Does this seem to be a valid estimate if, in a random sample of 90 college students, 28 are found to ride bicycles to class? Use a 0.05 level of significance

In the publication Relief from Arthritis by Thorsons Publishers L.td. John E Croft claims that over \(40 \%\) of the sufferers from osteoarthritis received measurable relief from an ingredient produced by a particular species of mussel found off the eoust of New Zealand. To test this claim, the mussel extract is to be given to a group of 7 osteonrthritic patients. If or more of the patients roccive relief, we shall not re-ject the null hypothesis that \(p=0.4\); otherwise, we conclude that \(p<0.4\) (a) Evaluate a, assuming that \(p=0.4\). (b) Evaluate \(\beta\) for the alternative \(p=0.3\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.