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Jacqueline Loya, a statistics student, asked students with jobs how many times they went out to eat in the last week. There were 25 students who had part- time jobs, and 25 students who had full-time jobs. Carry out a hypothesis test to determine whether the mean number of meals out per week for students with full-time jobs is greater than that for those with part-time jobs. Use a significance level of \(0.05 .\) Assume that the conditions for a two-sample \(t\) -test hold. Full-time jobs: \(5,3,4,4,4,2,1,5,6,5,6,3,3,2,4,5,2,3,7,5,5\), \(1,4,6,7\) Part-time jobs: \(1,1,5,1,4,2,2,3,3,2,3,2,4,2,1,2,3,2,1,3,3\), \(2,4,2,1\)

Short Answer

Expert verified
Based on the statistical t-test results, if the test statistic is greater than the critical value, we reject the null hypothesis, suggesting that the mean number of meals out per week for students with full-time jobs is greater than that for those with part-time jobs. Otherwise, we fail to reject the null hypothesis, suggesting that the means are equal or we don't have enough evidence to indicate a difference.

Step by step solution

01

Calculate the sample means and sample standard deviations.

For both full-time and part-time job students, calculate the sample means and sample standard deviations. Let's denote \( \bar{x}_{FT} \) as the sample mean for full-time students, \( \bar{x}_{PT} \) as the sample mean for part-time students, \( s_{FT} \) as the sample standard deviation for full-time students, and \( s_{PT} \) as the sample standard deviation for part-time students.
02

Set up the hypotheses.

The null hypothesis, denoted \( H_0 \), is the statement that the means are equal, i.e. 'the mean number of meals out per week for students with full-time jobs is equal to that for those with part-time jobs'. The alternative hypothesis, denoted \(H_1\), is 'the mean number of meals out per week for students with full-time jobs is greater than that for those with part-time jobs'. Mathematically: \( H_0: \mu_{FT} = \mu_{PT} \) and \( H_1: \mu_{FT} > \mu_{PT} \)
03

Calculate the test statistic.

Using the formula for the t-value in the two-sample t-test \( t = \frac{\bar{x}_{FT} - \bar{x}_{PT}}{\sqrt{\frac{s_{FT}^{2}}{n_{FT}} + \frac{s_{PT}^{2}}{n_{PT}}} }\), where \( \bar{x}_{FT} \) and \( \bar{x}_{PT} \) are the sample means, \( s_{FT} \) and \( s_{PT} \) are the standard deviations, and \( n_{FT} \) and \( n_{PT} \) are the sample sizes.
04

Calculate the degrees of freedom and critical value.

Calculate the degrees of freedom using the following formula for two independent samples: \( df = n_{FT} + n_{PT} - 2 \). Use the t-distribution table to find the critical t value for the significance level \( \alpha = 0.05 \) and the calculated degrees of freedom.
05

Make a decision regarding the null hypothesis.

Depending on whether the test-statistic is higher than the critical value, the decision is either to reject or not to reject the null hypothesis. If t > critical value, reject \( H_0 \) in favor of \( H_1 \).

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Key Concepts

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

Two-sample t-test
A two-sample t-test is a statistical method used to compare the means of two independent groups to see if there is a significant difference between them. This test is handy when determining whether, on average, the groups show different outcomes based on a specific measure.
In our scenario, the two groups are students with full-time jobs and students with part-time jobs. We're interested in finding out whether the mean number of meals eaten out per week differs between these two groups.
Key steps in using a two-sample t-test include:
  • Calculating the mean and standard deviation for each group.
  • Setting up the null and alternative hypotheses.
  • Using the formula for the t-value: \( t = \frac{\bar{x}_{FT} - \bar{x}_{PT}}{\sqrt{\frac{s_{FT}^{2}}{n_{FT}} + \frac{s_{PT}^{2}}{n_{PT}}} } \), where \( \bar{x} \) represents the sample mean and \( s \) the standard deviation.
  • Comparing the calculated t-value against a critical value from the t-distribution table to make a statistical decision.
The two-sample t-test assumes that the data from each group are independent, normally distributed, and have equal variances. However, adjustments may be made if these assumptions are violated.
Statistical Significance
Statistical significance is about determining if the result of a test can be attributed to something other than random chance. In hypothesis testing, it is determined using a p-value or a critical value. You may recall the significance level, denoted by \( \alpha \), which serves as a threshold for making this determination, typically set at 0.05.When a test is statistically significant, it means there's strong evidence against the null hypothesis, pointing towards the effect being observed as real rather than due to chance. In the exercise, we use a significance level of 0.05 to test whether full-time working students eat out more than those working part-time.
  • If the calculated t-value exceeds the critical value, the result is statistically significant, leading to the rejection of the null hypothesis.
  • A result is significant when the test statistic falls within the critical region at the tail end of the probability distribution.
Remember, statistical significance does not measure the size of an effect or its practical significance; it merely suggests that one exists.
Null and Alternative Hypotheses
The null and alternative hypotheses form the basis of hypothesis testing, acting as opposing statements that enable statistical examination of assumptions. The null hypothesis (\( H_0 \)) posits that there is no effect or difference, serving as a default stance we aim to test against. In contrast, the alternative hypothesis (\( H_1 \)), is what researchers assume might be true.
In Jacqueline's hypothesis test:
  • The null hypothesis is that the mean number of meals out per week for students with full-time jobs is equal to that for part-time job students \( (H_0: \mu_{FT} = \mu_{PT}) \).
  • The alternative hypothesis suggests that students with full-time jobs eat out more frequently \( (H_1: \mu_{FT} > \mu_{PT}) \).
The goal is to use statistical measures to determine whether the data supports the null hypothesis. If the data indicate significant evidence against it, the null is rejected in favor of the alternative hypothesis.
In summary, a clear understanding of both hypotheses is crucial, as this sets the groundwork for meaningful statistical investigation.

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Most popular questions from this chapter

In the United States, the population mean height for 3 -year-old boys is 38 inches (http://www.kidsgrowth .com). Suppose a random sample of 15 non-U.S. 3 -year-old boys showed a sample mean of \(37.2\) inches with a standard deviation of 3 inches. The boys were independently sampled. Assume that heights are Normally distributed in the population. a. Determine whether the population mean for non-U.S. boys is significantly different from the U.S. population mean. Use a significance level of \(0.05\) b. Now suppose the sample consists of 30 boys instead of 15, and repeat the test. c. Explain why the \(t\) -values and \(\mathrm{p}\) -values for parts a and \(\mathrm{b}\) are different.

State whether each of the following changes would make a confidence interval wider or narrower. (Assume that nothing else changes.) a. Changing from a \(95 \%\) level of confidence to a \(90 \%\) level of confidence b. Changing from a sample size of 30 to a sample size of 20 c. Changing from a standard deviation of 3 inches to a standard deviation of \(2.5\) inches

Choose a \(t\) -test for each situation: one-sample \(t\) -test, twosample \(t\) -test, paired \(t\) -test, and no \(t\) -test. a. A random sample of car dealerships is obtained. Then a student walks onto each dealer's lot wearing old clothes and finds out how long it takes (in seconds) for a salesperson to approach the student. Later the student goes onto the same lot dressed very nicely and finds out how long it takes for a salesperson to approach. b. A researcher at a preschool selects a random sample of 4 -year-olds, determines whether they know the alphabet (yes or no), and records gender. c. A researcher calls the office phone for a random sample of faculty at a college late at night, measures the length of the outgoing message, and records gender.

A statistics instructor randomly selected four bags of oranges, each bag labeled 10 pounds, and weighed the bags. They weighed \(10.2,10.5,10.3\), and \(10.3\) pounds. Assume that the distribution of weights is Normal. Find a \(95 \%\) confidence interval for the mean weight of all bags of oranges. Use technology for your calculations. a. Decide whether each of the following three statements is a correctly worded interpretation of the confidence interval, and fill in the blanks for the correct option(s). i. I am \(95 \%\) confident that the population mean is between ____. ii. There is a \(95 \%\) chance that all intervals will be between ____. and ____. iii. I am \(95 \%\) confident that the sample mean is between____. and ____. b. Does the interval capture 10 pounds? Is there enough evidence to reject the null hypothesis that the population mean weight is 10 pounds? Explain your answer.

Assume women's heights are approximately Normally distributed with a mean of 65 inches and a standard deviation of \(2.5\) inches. Which of the following questions can be answered using the Central Limit Theorem for sample means as needed? If the question can be answered, do so. If the question cannot be answered, explain why the Central Limit Theorem cannot be applied. a. Find the probability that a randomly selected woman is less than 63 inches tall. b. If five women are randomly selected, find the probability that the mean height of the sample is less than 63 inches. c. If 30 women are randomly selected, find the probability that the mean height of the sample is less than 63 inches.

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