/*! 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 24 Large Data Set 1 lists the GPAs ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Large Data Set 1 lists the GPAs of 1,000 students. a. Regard the data as arising from a census of all freshman at a small college at the end of their first academic year of college study, in which the GPA of every such person was measured. Compute the population mean \(\mu\). b. Regard the first 50 students in the data set as a random sample drawn from the population of part (a) and use it to test the hypothesis that the population mean is less than \(2.50,\) at the \(10 \%\) level of significance. (The null hypothesis is that \(\mu=2.50 .)\) c. Is your conclusion in part \((b)\) in agreement with the true state of nature (which by part (a) you know), or is your decision in error? If your decision is in error, is it a Type I error or a Type II error?

Short Answer

Expert verified
Find the mean GPA of all 1,000 students and test if the sample supports \( \mu < 2.50 \). If it doesn't match the known mean, determine the error type.

Step by step solution

01

Calculate Population Mean (Part a)

To find the population mean \( \mu \), sum all GPAs from Large Data Set 1 and divide by 1,000 (the total number of students). Let \( X_1, X_2, \ldots, X_{1000} \) represent the GPAs of the students, then the population mean \( \mu \) is computed as:\[\mu = \frac{X_1 + X_2 + \cdots + X_{1000}}{1000}.\]
02

Set Up Hypothesis Test (Part b)

We are testing the hypothesis that the population mean is less than 2.50. This means:- Null Hypothesis \( H_0: \mu = 2.50 \)- Alternative Hypothesis \( H_a: \mu < 2.50 \).
03

Calculate Sample Statistics

From the first 50 GPAs (the sample), calculate the sample mean \( \bar{x} \) and standard deviation \( s \). Let these be denoted as \( X_1, X_2, \ldots, X_{50} \). Compute:\[\bar{x} = \frac{X_1 + X_2 + \cdots + X_{50}}{50}\]and estimate the sample standard deviation \( s \) using the formula:\[s = \sqrt{\frac{\sum_{i=1}^{50} (X_i - \bar{x})^2}{49}}\]
04

Determine Test Statistic and Critical Value

Calculate the test statistic using the formula for a one-sample t-test:\[t = \frac{\bar{x} - 2.50}{s / \sqrt{50}}.\]Determine the critical value for \( t \) at \( 49 \) degrees of freedom at the 10% significance level for a one-tailed test using a t-distribution table.
05

Make the Decision

Compare the test statistic with the critical value:- If the test statistic is less than the critical value, reject the null hypothesis \( H_0 \).- Otherwise, do not reject \( H_0 \).
06

Evaluate Conclusion (Part c)

Based on the decision in Step 5, compare to the actual population mean from Step 1:- If you rejected \( H_0 \) and \( \mu \) is really less than 2.50, your decision is correct.- If you did not reject \( H_0 \) but \( \mu \) is less than 2.50, your decision is a Type II error.- If you rejected \( H_0 \) and \( \mu \) actually equals 2.50, your decision is a Type I error.

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.

Understanding the Population Mean
The population mean is the true average of a particular characteristic in a complete population. In our exercise, the population consists of 1,000 students and their GPAs. To find the population mean \(\mu\), we sum all of the GPAs from this group and then divide by 1,000.

This value gives us a clearer picture of the general performance level of all freshmen at the end of their first year. Since this figure encompasses every individual in the population, it is not an estimate but an exact measurement of average GPA at that school.
Exploring the Sample Mean
When we don't have access to data for an entire population, we turn to a sample mean as an estimate. In our use case, the first 50 students from the 1,000 are treated as a sample. The sample mean \(\bar{x}\) is calculated by adding together the GPAs of these 50 students and then dividing by 50.

This sample mean serves as a proxy to help approximate the population mean. However, the sample mean is always an estimate and can vary depending on which group of students are selected.
Type I and Type II Errors Explained
When conducting a hypothesis test, errors can occur. A Type I error happens when a true null hypothesis is rejected. In the exercise, if we conclude that the population mean GPA is less than 2.50, but in reality, it is equal to 2.50, we've made a Type I error.

Conversely, a Type II error occurs when we fail to reject a false null hypothesis. This would mean we conclude the GPA is at or above 2.50, while in fact it is below 2.50. Understanding these errors is crucial since they influence the confidence level and potential consequences of the test results.
Diving into Critical Value and Test Statistic
In hypothesis testing, the test statistic determines if our sample data supports rejecting the null hypothesis. It's calculated using the sample mean, assumed population mean under null hypothesis, sample standard deviation, and sample size.

The critical value is derived from a statistical distribution (like t-distribution) at a specific significance level. In this context, it's the threshold that the test statistic is compared to. If the test statistic is more extreme than the critical value, the null hypothesis is rejected.

This comparison tells us whether there is enough evidence in our sample to infer that the population parameter is different from the hypothesized value.

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

An economist wishes to determine whether people are driving less than in the past. In one region of the country the number of miles driven per household per year in the past was 18.59 thousand miles. A sample of 15 households produced a sample mean of 16.23 thousand miles for the last year, with sample standard deviation 4.06 thousand miles. Assuming a normal distribution of household driving distances per year, perform the relevant test at the \(5 \%\) level of significance.

A dairy farm uses the somatic cell count (SCC) report on the milk it provides to a processor as one way to monitor the health of its herd. The mean SCC from five samples of raw milk was 250,000 cells per milliliter with standard deviation 37,500 cell/ml. Test whether these data provide sufficient evidence, at the \(10 \%\) level of significance, to conclude that the mean SCC of all milk produced at the dairy exceeds that in the previous report, 210,250 cell/ml. Assume a normal distribution of SCC.

The mean household income in a region served by a chain of clothing stores is $$\$ 48,750 .$$ In a sample of 40 customers taken at various stores the mean income of the customers was $$\$ 51,505$$ with standard deviation $$\$ 6,852 .$$ a. Test at the \(10 \%\) level of significance the null hypothesis that the mean household income of customers of the chain is $$\$ 48,750$$ against that alternative that it is different from $$\$ 48,750.$$ b. The sample mean is greater than $$\$ 48,750,$$ suggesting that the actual mean of people who patronize this store is greater than $$\$ 48,750 .$$ Perform this test, also at the \(10 \%\) level of significance. (The computation of the test statistic done in part (a) still applies here.)

Find the rejection region (for the standardized test statistic) for each hypothesis test. Identify the test as left-tailed, right-tailed, or two- tailed. a. \(\quad H 0: \mu=141\) VS. Ha: \(\mu<141\) \(@ \alpha=0.20\). b. \(\quad H 0: \mu=-54\) vs. Ha: \(\mu<-54 @ \alpha=0.05\). C. \(\quad H 0: \mu=98.6\) VS. \(H a: \mu \neq 98.6 @ \alpha=0.05 .\) d. \(\quad H 0: \mu=3.8\) VS. Ha: \(\mu>3.8 @ \alpha=0.001\).

At every setting a high-speed packing machine delivers a product in amounts that vary from container to container with a normal distribution of standard deviation 0.12 ounce. To compare the amount delivered at the current setting to the desired amount 64.1 ounce, a quality inspector randomly selects five containers and measures the contents of each, obtaining sample mean 63.9 ounces and sample standard deviation 0.10 ounce. Test whether the data provide sufficient evidence, at the \(5 \%\) level of significance, to conclude that the mean of all containers at the current setting is less than 64.1 ounces.

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.