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Chapter 10: Problem 79

A student at a four-year college claims that mean enrollment at four鈥搚ear colleges is higher than at two鈥搚ear colleges in the United States. Two surveys are conducted. Of the 35 two鈥搚ear colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191.

Short Answer

Expert verified
No statistical evidence supports higher mean enrollment at four-year colleges.

Step by step solution

01

Defining Hypotheses

We want to test the claim that the mean enrollment of four-year colleges (\(ar{x}_2 = 5466\)) is greater than the mean enrollment of two-year colleges (\(ar{x}_1 = 5068\)). The null hypothesis\(H_0\) is: the mean enrollment for four-year and two-year colleges is equal. The alternative hypothesis\(H_1\) is: the mean enrollment for four-year colleges is greater than two-year colleges.
02

Determine Sample Characteristics

We have two samples: For two-year colleges, the sample mean\(ar{x}_1 = 5068\), standard deviation\(s_1 = 4777\), and sample size\(n_1 = 35\). For four-year colleges, the sample mean\(ar{x}_2 = 5466\), standard deviation\(s_2 = 8191\), and sample size\(n_2 = 35\).
03

Conduct a t-Test for Difference in Means

We use a two-sample t-test with unequal variances (Welch's t-test) because the standard deviations differ. Calculate the t-statistic using the formula:\[t = \frac{\bar{x}_2 - \bar{x}_1}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\] Substitute the values:\[t = \frac{5466 - 5068}{\sqrt{\frac{4777^2}{35} + \frac{8191^2}{35}}}\]Calculate the numerator, the denominator, and then the t-value.
04

Calculate the t-Statistic

First calculate the numerator: \(5466 - 5068 = 398\).Now the denominator: \(\sqrt{\frac{4777^2}{35} + \frac{8191^2}{35}}\).Calculate:\(\frac{4777^2}{35} \approx 650544.57\)and\(\frac{8191^2}{35} \approx 1916624.39\).Add these:\(650544.57 + 1916624.39 = 2567168.96\), then \(\sqrt{2567168.96} \approx 1601.38\).Thus, the t-statistic is:\(\frac{398}{1601.38} \approx 0.248\).
05

Determine Significance

With \(df\) degrees of freedom calculated using Welch's approximation,approximate using sample sizes and variances (\(df \approx 67.86\)).Look up \(t\)-critical for one-tailed test at the desired significance level, (e.g., 0.05) for \(df = 67.86\).Find \(t\)-critical approximately (typically about 1.67).Compare this with calculated \(t\)-statistic (0.248).
06

Conclusion and Decision

Since the calculated \(t\)-statistic (0.248) is less than the \(t\)-critical value (1.67), we do not reject the null hypothesis. Therefore, there is insufficient evidence to support the claim that mean enrollment is higher at four-year colleges compared to two-year colleges.

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Key Concepts

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

t-test
A t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It helps us understand whether any observed difference could have happened by chance.
When we have two independent samples, such as enrollments from two-year and four-year colleges, we can use a t-test to see if there's a real difference in their average sizes.
  • Independent Samples: Ensure that the two groups are independent, meaning what's happening in one group shouldn't influence the other.
  • Type of t-test: In scenarios where the variances aren't assumed to be equal (like our exercise), Welch's t-test is appropriate.
In summary, the t-test is a powerful tool for hypothesis testing, especially when we want to compare averages from two groups.
sample statistics
Sample statistics are numbers that summarize data from a sample. In our exercise, we have two groups, each with their own mean and standard deviation. These figures help us test hypotheses about larger populations.
Here's how it breaks down:
  • Mean (\(ar{x}\)): This is the average enrollment, calculated by summing up all students in a group and dividing by the number of colleges surveyed.
  • Standard Deviation (\(s\)): This measures the variation or dispersion from the average enrollment. A higher standard deviation indicates students numbers are spread out more widely from the mean.
  • Sample Size (\(n\)): The number of colleges surveyed in each group. A larger sample size generally provides more reliable statistics.
These statistics form the basis for conducting our t-test. They help us determine differences between the two samples, which can reveal patterns about the wider population.
Welch's t-test
Welch's t-test is a version of the t-test used when we have two samples with unequal variances.
In our exercise, the two-year and four-year colleges have quite different standard deviations. This makes Welch's t-test the best choice because it doesn't assume equal variances.
Here's why Welch's t-test is suitable:
  • No Equal Variance Assumption: Unlike the standard t-test, Welch's t-test does not assume that both groups have the same variance, making it adaptable.
  • Use in Real-world Situations: This test is often used in practical scenarios where equal variances aren't realistic, like our college survey.
  • Calculation: It involves the mean, standard deviation, and sample size of both groups, providing a more flexible approach.
Welch鈥檚 t-test is particularly helpful when you want accurate results for two samples with differing variance levels.
null and alternative hypotheses
Null and alternative hypotheses form the backbone of hypothesis testing in statistics.
They provide a structured way to test assumptions or claims about the population using sample data. Here's how they are defined in our exercise:
  • Null Hypothesis (\(H_0\)): This statement suggests that there is no difference between the mean enrollments of four-year and two-year colleges. It's the 'status quo' assertion that we test against.
  • Alternative Hypothesis (\(H_1\)): This hypothesis posits that the mean enrollment of four-year colleges is greater than that of two-year colleges. It's what we suspect might be true, given our initial claim.
In hypothesis testing, we use statistical tests (like the t-test) to determine whether there is enough evidence to reject the null hypothesis in favor of the alternative hypothesis.

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Most popular questions from this chapter

Joan Nguyen recently claimed that the proportion of college-age males with at least one pierced ear is as high as the proportion of college-age females. She conducted a survey in her classes. Out of 107 males, 20 had at least one pierced ear. Out of 92 females, 47 had at least one pierced ear. Do you believe t at the proportion of males has reached the proportion of females?

Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. a. independent group means, population standard deviations and/or variances known b. independent group means, population standard deviations and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it.

Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites. Is this a test of means or proportions?

Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. $$\begin{array}{|l|l|l|}\hline & {\text { Sample Mean Melting Temperatures }\left(^{\circ} F\right)} & {\text { Population Standard Deviation }} \\\ \hline \text { Alloy Gamma } & {800}&{95} \\ \hline \text { Alloy zeta } & {900} &{105} \\ \hline\end{array}$$ What is the p-value?

Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. a. independent group means, population standard deviations and/or variances known b. independent group means, population standard deviations and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and one year, respectively.
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