/*! 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 8 Women athletes at the University... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 8

Women athletes at the University of Colorado, Boulder, have a long-term graduation rate of \(67 \%\) (Source: The Cbronicle of Higher Education). Over the past several years, a random sample of 38 women athletes at the school showed that 21 eventually graduated. Does this indicate that the population proportion of women athletes who graduate from the University of Colorado, Boulder, is now less than \(67 \%\) ? Use a \(5 \%\) level of significance.

Short Answer

Expert verified
No, there is not enough evidence to suggest the graduation rate is less than \(67\%\) at a \(5\%\) significance level.

Step by step solution

01

Define the Hypotheses

We set up our null and alternative hypotheses. The null hypothesis \(H_0\) is that the population proportion \(p\) of women athletes who graduate is \(67\%\). The alternative hypothesis \(H_a\) is that the population proportion is less than \(67\%\). Thus, we have:\[ H_0: p = 0.67 \]\[ H_a: p < 0.67 \]
02

Collect Sample Data

From the problem, we have a sample size \(n = 38\) and the number of women athletes who graduated is \(x = 21\). The sample proportion \(\hat{p}\) is calculated as:\[ \hat{p} = \frac{x}{n} = \frac{21}{38} \approx 0.5526 \]
03

Calculate the Test Statistic

We will use the formula for the test statistic for a proportion:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]Plugging in the values:\[ z = \frac{0.5526 - 0.67}{\sqrt{\frac{0.67(1-0.67)}{38}}} \]Calculate the denominator:\[ \sqrt{\frac{0.67 \times 0.33}{38}} \approx 0.0775 \]Then the test statistic is:\[ z \approx \frac{-0.1174}{0.0775} \approx -1.514 \]
04

Determine the Critical Value and P-Value

For a \(5\%\) level of significance in a left-tailed test, we find the critical value from the standard normal distribution table, which is approximately \(-1.645\). We look up the p-value for \(z \approx -1.514\), which is about \(0.065\).
05

Make a Decision

Since the p-value \(0.065 > 0.05\) and the test statistic \(-1.514\) is greater than the critical value \(-1.645\), we do not reject the null hypothesis. There is not enough evidence to conclude that the proportion of women athletes who graduate has decreased to less than \(67\%\).

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 91影视!

Key Concepts

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

Sample Proportion
To understand hypothesis testing, it's crucial to grasp the concept of sample proportion. In sampling, the sample proportion (\(\hat{p}\)) acts as an estimate of the true population proportion. Put simply, it's the fraction of items in our sample that possess a particular attribute.
In our exercise, the sample proportion is the fraction of women athletes who graduated from the sample group. With 21 out of 38 athletes graduating, the sample proportion calculated is:\[ \hat{p} = \frac{21}{38} \approx 0.5526 \]This calculation tells us that roughly 55.26% of the sampled women athletes graduated.
When testing hypotheses, this sample result is compared to a hypothesized population proportion to determine if there's a significant difference.
Test Statistic
The test statistic is a crucial component in hypothesis testing. It uses sample data to determine how far away the sample proportion is from the hypothesized population proportion. The standard test statistic for proportion is represented by the standard normal distribution (z-distribution).
For our scenario, the test statistic (\(z\)) is computed as follows:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]Here, \(\hat{p}\) is our sample proportion, \(p_0\) is the hypothesized population proportion (67% in this case), and \(n\)is the sample size.
In our example, the calculated test statistic is approximately \(z \approx -1.514\), indicating the number of standard deviations our sample proportion is from the hypothesized value.
Null Hypothesis
The null hypothesis (\(H_0\)) is the foundation of hypothesis testing. It assumes no effect or difference in the population parameter unless evidence suggests otherwise.
In our exercise, the null hypothesis states that the proportion (\(p\)) of women athletes who graduate is equal to 67%. Formally, we write:\[ H_0: p = 0.67 \]The null hypothesis provides a baseline that we test against. Our goal is to determine whether there's enough statistical evidence to reject this assumption in favor of an alternative.
Alternative Hypothesis
The alternative hypothesis (\(H_a\)) presents a contrasting claim to the null hypothesis. It suggests that a specific effect or difference might exist.
In our scenario, the alternative hypothesis posits that the proportion (\(p\)) of women athletes who graduate is actually less than 67%. This is expressed as:\[ H_a: p < 0.67 \]
If our statistical analysis yields enough evidence against the null hypothesis, we would accept the alternative hypothesis. However, in this exercise, since the p-value (0.065) is greater than our significance level (0.05), we do not have sufficient evidence to support the alternative hypothesis.

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

Is fishing better from a boat or from the shore? Pyramid Lake is located on the Paiute Indian Reservation in Nevada. Presidents, movie stars, and people who just want to catch fish go to Pyramid Lake for really large cutthroat trout. Let row \(B\) represent hours per fish caught fishing from the shore, and let row \(A\) represent hours per fish caught using a boat. The following data are paired by month from October through April. (Source: Pyramid Lake Fisheries, Paiute Reservation, Nevada.) \(\begin{array}{l|ccccccc} \hline & \text { Oct. } & \text { Nov. } & \text { Dec. } & \text { Jan. } & \text { Feb. } & \text { March } & \text { April } \\ \hline \text { B: Shore } & 1.6 & 1.8 & 2.0 & 3.2 & 3.9 & 3.6 & 3.3 \\ \hline \text { A: Boat } & 1.5 & 1.4 & 1.6 & 2.2 & 3.3 & 3.0 & 3.8 \\ \hline \end{array}\) Use a \(1 \%\) level of significance to test if there is a difference in the population mean hours per fish caught using a boat compared with fishing from the shore.

Let \(x\) be a random variable that represents the \(\mathrm{pH}\) of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the \(x\) distribution is \(\mu=7.4\) (Reference: Merck Manual, a commonly used reference in medical schools and nursing programs). A new drug for arthritis has been developed. However, it is thought that this drug may change blood \(\mathrm{pH} . \mathrm{A}\) random sample of 31 patients with arthritis took the drug for 3 months. Blood tests showed that \(\bar{x}=8.1\) with sample standard deviation \(s=1.9 .\) Use a \(5 \%\) level of significance to test the claim that the drug has changed (either way) the mean \(\mathrm{pH}\) level of the blood.

In environmental studies, sex ratios are of great importance. Wolf society, packs, and ecology have been studied extensively at different locations in the U.S. and foreign countries. Sex ratios for eight study sites in northern Europe are shown below (based on The Wolf by L. D. Mech, University of Minnesota Press). \(\begin{array}{lcc} \hline \text { Location of Wolf Pack } & \text { \% Males (Winter) } & \text { \% Males (Summer) } \\ \hline \text { Finland } & 72 & 53 \\ \text { Finland } & 47 & 51 \\ \text { Finland } & 89 & 72 \\ \text { Lapland } & 55 & 48 \\ \text { Lapland } & 64 & 55 \\ \text { Russia } & 50 & 50 \\ \text { Russia } & 41 & 50 \\ \text { Russia } & 55 & 45 \\ \hline \end{array}\) It is hypothesized that in winter, "loner" males (not present in summer packs) join the pack to increase survival rate. Use a \(5 \%\) level of significance to test the claim that the average percentage of males in a wolf pack is higher in winter.

USA Today reported that the state with the longest mean life span is Hawaii, where the population mean life span is 77 years. A random sample of 20 obituary notices in the Honolulu Advertizer gave the following information about life span (in years) of Honolulu residents: \(\begin{array}{llllllllll} 72 & 68 & 81 & 93 & 56 & 19 & 78 & 94 & 83 & 84 \\ 77 & 69 & 85 & 97 & 75 & 71 & 86 & 47 & 66 & 27 \end{array}\) i. Use a calculator with mean and standard deviation keys to verify that \(\bar{x}=\) \(71.4\) years and \(s\) \& \(20.65\) years. ii. Assuming that life span in Honolulu is approximately normally distributed, does this information indicate that the population mean life span for Honolulu residents is less than 77 years? Use a \(5 \%\) level of significance.

REM (rapid eye movement) sleep is sleep during which most dreams occur. Each night a person has both REM and non-REM sleep. However, it is thought that children have more REM sleep than adults (Reference: Secrets of Sleep by Dr. A. Borbely). Assume that REM sleep time is normally distributed for both children and adults. A random sample of \(n_{1}=10\) children ( 9 years old) showed that they had an average REM sleep time of \(\bar{x}_{1}=2.8\) hours per night. From previous studies, it is known that \(\sigma_{1}=0.5\) hour. Another random sample of \(n_{2}=10\) adults showed that they had an average REM sleep time of \(\bar{x}_{2}=2.1\) hours per night. Previous studies show that \(\sigma_{2}=\) \(0.7\) hour. Do these data indicate that, on average, children tend to have more REM sleep than adults? Use a \(1 \%\) level of significance.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation
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.