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Chapter 4: Problem 13

A situation is described for a statistical test. In each case, define the relevant parameter(s) and state the null and alternative hypotheses. Testing to see if there is evidence that the mean time spent studying per week is different between first-year students and upperclass students

Short Answer

Expert verified
The relevant parameters are the mean time spent studying per week for first-year students (\( \mu_{1} \)) and upperclass students (\( \mu_{2} \)). The null hypothesis is \( H0: \mu_{1} - \mu_{2} = 0 \), meaning there is no difference in the mean study time between the two groups. The alternative hypothesis is \( H1: \mu_{1} - \mu_{2} \neq 0 \), indicating that there is a difference in the mean study time between the two groups.

Step by step solution

01

Identify the parameters

The parameters for this case are the mean time studying per week for first-year students (\( \mu_{1} \)) and for upperclass students (\( \mu_{2} \)).
02

State the Null Hypothesis

The Null Hypothesis (H0) is that there is no difference in the mean time spent studying between the two groups. In mathematical terms, this can be represented as: \( H0: \mu_{1} - \mu_{2} = 0 \)
03

State the Alternative Hypothesis

The Alternative Hypothesis (H1) is that there is a significant difference between the mean time spent studying between the two groups. Mathematically, this can be represented as: \( H1: \mu_{1} - \mu_{2} \neq 0 \)

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Key Concepts

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

Null Hypothesis
When conducting a hypothesis test, we begin with the null hypothesis. The null hypothesis is a statement that suggests there is no effect or no difference, and it serves as a starting point for statistical testing. In our case, we want to check if there is any difference between the mean time spent studying by first-year students compared to upperclass students.
The null hypothesis (\( H_0 \)) formulaically suggests:
  • There is no difference in the mean study time (\( \mu_1 - \mu_2 = 0 \));
Here, \( \mu_1 \) is the mean study time for first-year students and \( \mu_2 \) is the mean study time for upperclass students. If the data supports the null hypothesis, it means any observed difference in means might be due to random chance.
Essentially, the null hypothesis sets the standard we need to disprove with our data to show any statistically significant difference exists between the two groups.
Alternative Hypothesis
Contrary to the null hypothesis, the alternative hypothesis posits that there is, in fact, a difference or effect. It serves as the claim we are trying to find evidence for in our hypothesis test. In this exercise, we're interested in determining whether the difference in mean study times between first-year and upperclass students is statistically significant.
The alternative hypothesis (\( H_1 \)) can be stated as:
  • There is a significant difference in the mean study time (\( \mu_1 - \mu_2 eq 0 \));
This means if our test provides enough evidence to reject the null hypothesis, we accept the alternative, indicating a real difference in study habits between the two student groups. The direction or extent of the difference isn't specified here, just that a difference exists. This is why it's considered two-sided; we're open to first-year students studying more or less than their upperclass peers.
Mean Comparison
In hypothesis testing, comparison of means is a crucial aspect when we wish to determine if two groups differ in a meaningful way. For our particular context, we're comparing the average study times of first-year students and upperclass students.
This comparison of means can reveal whether the observed data is likely or unlikely under the assumption that the null hypothesis is true. When the difference in means is large enough, it may indicate a significant finding that supports the alternative hypothesis.
  • We compute the difference between the means of two groups: \(\bar{x}_1 - \bar{x}_2\)
  • Apply a statistical test (like a t-test) to determine if observed differences are statistically significant;
In essence, mean comparison is about examining whether the average values of two samples give us enough evidence to draw conclusions about the populations they represent. This analysis is fundamental to validating claims about different population behaviors based on sampled data.

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Most popular questions from this chapter

Classroom Games Exercise 4.62 describes a situation in which game theory students are randomly assigned to play either Game 1 or Game 2 , and then are given an exam containing questions on both games. Two one-tailed tests were conducted: one testing whether students who played Game 1 did better than students who played Game 2 on the question about Game \(1,\) and one testing whether students who played Game 2 did better than students who played Game 1 on the question about Game \(2 .\) The p-values were 0.762 and 0.549 , respectively. The p-values greater than 0.5 mean that, in the sample, the students who played the opposite game did better on each question. What does this study tell us about possible effects of actually playing a game and answering a theoretical question about it? Explain.

In Exercises 4.112 to \(4.116,\) the null and alternative hypotheses for a test are given as well as some information about the actual sample(s) and the statistic that is computed for each randomization sample. Indicate where the randomization distribution will be centered. In addition, indicate whether the test is a left-tail test, a right-tail test, or a twotailed test. Hypotheses: \(H_{0}: \mu=10\) vs \(H_{a}: \mu>10\) Sample: \(\bar{x}=12, s=3.8, n=40\) Randomization statistic \(=\bar{x}\)

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Car Window Skin Cancer? A new study suggests that exposure to UV rays through the car window may increase the risk of skin cancer. \(^{43}\) The study reviewed the records of all 1050 skin cancer patients referred to the St. Louis University Cancer Center in 2004 . Of the 42 patients with melanoma, the cancer occurred on the left side of the body in 31 patients and on the right side in the other 11 . (a) Is this an experiment or an observational study? (b) Of the patients with melanoma, what proportion had the cancer on the left side? (c) A bootstrap \(95 \%\) confidence interval for the proportion of melanomas occurring on the left is 0.579 to \(0.861 .\) Clearly interpret the confidence interval in the context of the problem. (d) Suppose the question of interest is whether melanomas are more likely to occur on the left side than on the right. State the null and alternative hypotheses. (e) Is this a one-tailed or two-tailed test? (f) Use the confidence interval given in part (c) to predict the results of the hypothesis test in part (d). Explain your reasoning. (g) A randomization distribution gives the p-value as 0.003 for testing the hypotheses given in part (d). What is the conclusion of the test in the context of this study? (h) The authors hypothesize that skin cancers are more prevalent on the left because of the sunlight coming in through car windows. (Windows protect against UVB rays but not UVA rays.) Do the data in this study support a conclusion that more melanomas occur on the left side because of increased exposure to sunlight on that side for drivers?

In Exercises 4.150 to \(4.152,\) a confidence interval for a sample is given, followed by several hypotheses to test using that sample. In each case, use the confidence interval to give a conclusion of the test (if possible) and also state the significance level you are using. A \(90 \%\) confidence interval for \(p_{1}-p_{2}: 0.07\) to 0.18 (a) \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1} \neq p_{2}\) (b) \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}>p_{2}\) (c) \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}
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