/*! 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 32 Amslat Neus (December 2004) list... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 32

Amslat Neus (December 2004) lists median salaries for associate professors of statistics at research institutions and at liberal arts and other institutions in the United States. Assume a sample of 200 associate professors from research institutions having an average salary of \(\$ 70.750\) per year with a standard deviation of S6000. Assume also a sample of 200 associate professors from other types of institutions having an average salary of \(\$ 65,200\) with a standard deviation of \(\$ 5000\) Test the hypothesis that the mean salary for associate professors in research institutions is \(\$ 2000\) higher than for those in other institutions. Use a 0.01 level of significance

Short Answer

Expert verified
The conclusion will be based on comparing the calculated test statistic with the critical value. A decision to reject or not reject the null hypothesis can be drawn from this comparison.

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis (\(H_0\)) states that the mean difference in salary between the research and other institutions is \$2000. The alternative hypothesis (\(H_1\)) is that the mean difference in salary is not \$2000.
02

Calculate the Test Statistic

The test statistic is calculated using the formula for two independent samples: \[\dfrac{(\bar{X_1} - \bar{X_2}) - D_0}{\sqrt{\dfrac{(S_1^2/n_1) + (S_2^2/n_2)}}}\]where \(\bar{X_1}\) and \(\bar{X_2}\) are the sample means, \(S_1\) and \(S_2\) are the sample standard deviations, \(n_1\) and \(n_2\) are the sample sizes, and \(D_0\) is the hypothesized difference (in this case, \$2000).
03

Compute the Test Statistic

Substitute the provided values into the formula: \[\dfrac{(70,750-65,200) - 2000}{\sqrt{\dfrac{(6000^2/200) + (5000^2/200)}}}\]
04

Determine the Critical Value

The critical value can be found using a Z-table. Given the 0.01 level of significance for a two-tailed test, the critical value (z) is approximately 卤2.58.
05

Decision Rule

If the computed test statistic is greater than the critical value in absolute terms, reject the null hypothesis. If not, fail to reject null hypothesis.
06

Compare and Make Conclusion

Compare the calculated test statistic with the critical value. Depending on the results, make a conclusion about whether the mean salary for associate professors in research institutions is \$2000 higher than for those in other institutions.

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.

Null Hypothesis
In statistics, the null hypothesis (ull hypothesis or ull hypothesis) serves as a starting point for hypothesis testing. It is a statement of no effect or no difference, and in the context of our exercise, it posits that the average salary difference between associate professors at research institutions and at other institutions is exactly \(2000.

To test this hypothesis, we perform a statistical test which could either lead us to reject the null hypothesis if the evidence suggests a difference other than \)2000, or fail to reject the null hypothesis if we do not find strong enough evidence to suggest otherwise.
Alternative Hypothesis
Contrary to the null hypothesis, the alternative hypothesis (ull hypothesis) is what a researcher aims to support. It suggests that there is an effect or a difference, and in this scenario, it states that the mean salary difference is not \(2000.

This is not a claim that the difference is more or less than \)2000; rather, it simply indicates that the difference is not equal to \(2000. If evidence indicates a mean salary difference other than \)2000, the alternative hypothesis gains credibility.
Test Statistic Calculation
The test statistic is a key element in hypothesis testing as it allows researchers to determine how far the sample statistic is from the null hypothesis. It is calculated using a specific formula that incorporates the sample mean, the hypothesized mean difference, and the sample standard deviation.

In our example, the formula provided calculates the Z-score which is a measure of how many standard deviations our sample mean is away from the hypothesized mean. It is paramount that students follow the formula accurately and understand how each component affects the calculation.
Level of Significance
The level of significance, denoted by ull hypothesis, is the threshold for determining whether a statistical result is significant. It represents the probability of rejecting the null hypothesis when it is true. A typical level of significance used is 0.05, but in more rigorous studies, a level of 0.01 or even 0.001 might be employed.

In the given exercise, a 0.01 level of significance is specified, implying that there is only a 1% chance we are willing to take to incorrectly reject the null hypothesis. Determining the level of significance before conducting the hypothesis test is critical, as it defines the benchmarks for making a decision regarding the hypotheses.
Two-Sample Z-Test
When dealing with two independent samples and we want to compare their means, we use a two-sample Z-test. This statistical test is specifically employed when we have large sample sizes (generally over 30) and known population standard deviations. The Z-test calculates the Z-score which helps us understand if the difference between the two sample means is statistically significant or not.

The Z-score is then compared to the critical value that corresponds to the chosen level of significance. If the absolute value of the Z-score is greater than the critical value from the Z-table, we reject the null hypothesis. If not, we fail to reject it. This test is particularly relevant to the example at hand, highlighting the importance of understanding how to calculate and interpret the results of a two-sample Z-test.

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

Past experience indicates that the time for high school seniors to complete a standardized test is a normal random variable with a mean of 35 minutes. If a random sample of 20 high school seniors took an average of 33.1 minutes to complete this test with a standard deviation of 4.3 minutes, test the hypothesis at the 0.05 level of significance that \(p=35\) minutes against the alternative that \(p<35\) minutes.

Past expericnce indicates that the time required for high school seniors to complete a standardized test is a normal random variable with a standard deviation of 6 minutes. Test the hypothesis that \(0=0\) against the alternative that \(1 \mathrm{~T}<6\) if a random sample of 20 high school seniors has a standard deviations \(s=4.51 .\) Use a 0.05 level of significance.

10.36 \text { A large automobile manufacturing company is } trying to decide whether to purchase brand \(A\) or brand \(B\) tires for its new models. To help arrive at a decision, an experiment is conducted using 12 of each brand. The tires are run until they wear out. The results are $$\begin{aligned}\text { Brand } A: & x_{1}=37,900 \text { kilometers, } \\ & s_{1}=5,100 \text { kilometers. } \\\\\text { Brand B: } \quad x_{1} &=39,800 \text { kilometers, } \\\& s_{2}=5,900 \text { kilometers. }\end{aligned}$$ Test the hypothesis that there is no difference in the average wear of 2 brands of tires. Assume the populations to be approximately normally distributed with equal variances. Use a P-value.

Large-Sample Test of \(a^{2}=\sigma_{0}^{2}\). When \(n \geq\) 30 we can test the null hypothesis that \(\sigma^{2}=a_{5}^{2}\) or \(\sigma-\) (Ta, by computing $$z=\frac{s+\sigma_{0}}{\sigma_{0} / \sqrt{2 n}}$$ which is a value of a random variable whose sampling distribution is approximately the standard normal distribution (a) With reference to Example 10.5 , test at the 0.05 level of significance whether \(\sigma=10.0\) years against the alternative that \(\sigma \neq 10.0\) years. (b) It is suspected that the variance of the distribution of distances in kilometers achieved per 5 liters of fuel by a new automobile model equipped with a diesel engine is less than the variance of the distribution of distances achieved by the same model equipped with a six-cylinder gasoline engine, which is known to be \(\sigma^{2}=6.25\). If 72 test runs in the diesel model have a variance of 4.41 , ean we conclude at the 0.05 level of significance that the variance of the distances achicved by the: diesel model is less than that of the gasoline model?

Three cards are drawn from an ordinary deck of playing cards, with replacement, and the number \(Y\) of spades is recorded. After repeating the experiment 64 times, the following outcomes were recorded: $$\begin{array}{c|cccc}y & 0 & 1 & 2 & 3 \\\\\hline \boldsymbol{f} & 21 & 31 & 12 & 0\end{array}$$ Test the hypothesis of 0.01 level of significance that the recorded data may be fitted by the binomial distribution \(b(y ; 3,1 / 4) y=0,1,2,3\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Statistics

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.