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Chapter 6: Problem 253

Test \(H_{0}: \mu_{T}=\mu_{C}\) vs \(H_{a}: \mu_{T}<\mu_{C}\) using the fact that the treatment group \((T)\) has a sample mean of 8.6 with a standard deviation of 4.1 while the control group \((C)\) has a sample mean of 11.2 with a standard deviation of \(3.4 .\) Both groups have 25 cases.

Short Answer

Expert verified
Based on the given data and the computed test statistic, we reject the null hypothesis at the 5% significance level and conclude that the treatment group mean is statistically significantly less than the control group mean.

Step by step solution

01

State the Hypotheses

The null hypothesis is \(H_{0}: \mu_{T}=\mu_{C}\) and the alternative hypothesis is \(H_{a}: \mu_{T}<\mu_{C}\). The null hypothesis states that the mean difference between treatment and control group is zero, while the alternative hypothesis states that the mean of the treatment group is less than the mean of the control group.
02

Calculate the Test Statistic

The test statistic for the two-sample mean comparison is computed as the difference in sample means, minus the difference under null hypothesis, divided by standard error of the difference. Given the data, it is \[z = \frac{{(\bar{x}_{T} - \bar{x}_{C}) - 0}}{{\sqrt{(\frac{{s_{T}^2}}{{n_{T}}}) + (\frac{{s_{C}^2}}{{n_{C}}}})} = \frac{{(8.6 - 11.2) - 0}}{{\sqrt{(\frac{{4.1^2}}{{25}}) + (\frac{{3.4^2}}{{25}}}} = -2.69.\]
03

Determine the Critical Value and Make a Decision

For the given hypotheses, this is a one-tail test and since the standard normal distribution is symmetric, the critical value for the 5% significance level is -1.645. Comparing the test statistic to the critical value, -2.69 < -1.645, we reject the null hypothesis.

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Key Concepts

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

Two-sample Mean Comparison
When comparing two samples, such as treatment and control groups, we want to determine if their means are significantly different. The two-sample mean comparison test helps us do just that. In our scenario, the treatment group has a sample mean of 8.6 and the control group has a sample mean of 11.2. These figures indicate the central tendency of each group.
This test is particularly useful when you're looking to compare two independent groups, especially in experimental settings or observational studies.
- **Null Hypothesis ( H_{0} )** assumes that both group means are equal. For example, you predict that a new teaching method (treatment group) is as effective as the traditional method (control group). - **Alternative Hypothesis ( H_{a} )** suggests that there's a difference in means. In this exercise, we hypothesize that the treatment group's mean is lower. By calculating and comparing means, we're able to make an informed decision about the effectiveness or impact between the groups.
Test Statistic
The test statistic is a crucial component in hypothesis testing. It provides a standardized way to compare observed data against the null hypothesis. In the case of a two-sample mean comparison, the test statistic is calculated using the formula: \[ z = \frac{(\bar{x}_{T} - \bar{x}_{C}) - 0}{\sqrt{(\frac{s_{T}^2}{n_{T}}) + (\frac{s_{C}^2}{n_{C}})}}\]The test statistic, denoted as "z" in this case, is derived from the difference in sample means and the standard error of the difference.
- **Sample Mean Difference:** Represents the difference between the means of the treatment and control groups.- **Standard Error (SE):** Measures how much the sample means are expected to vary. It is influenced by the standard deviation and sample size of both groups.In our example, with a test statistic of -2.69, we gain a measured value that helps in determining how unusual our observational data is if the null hypothesis were true.
Critical Value
The critical value is a threshold that helps determine whether the test statistic falls in the rejection region of the null hypothesis. It is determined based on the chosen significance level (often denoted as \(\alpha\)). In many cases, including ours, a common significance level is 5% or 0.05 for one-tailed tests.
The critical value (-1.645 in this example) represents the point beyond which we consider results significantly different enough to reject the null hypothesis. For our standard normal distribution:
  • The test is one-tailed since we're specifically testing if the treatment group's mean is less than the control group's.
  • A significance level of 0.05 implies a 5% risk of concluding a difference exists when there is none.
  • If our test statistic is less than the critical value, we reject the null hypothesis.
With a test statistic of -2.69, which is less than the critical value of -1.645, there is enough evidence to reject the null hypothesis and support the claim that the treatment group's mean is indeed lower.

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Most popular questions from this chapter

THC vs Prochloroperazine An article in the New York Times on January 17,1980 reported on the results of an experiment that compared an existing treatment drug (prochloroperazine) with using THC (the active ingredient in marijuana) for combating nausea in patients undergoing chemotherapy for cancer. Patients being treated in a cancer clinic were divided at random into two groups which were then assigned to one of the two drugs (so they did a randomized, double- blind, comparative experiment). Table 6.15 shows how many patients in each group found the treatment to be effective or not effective. (a) Use these results to test whether the proportion of patients helped by THC is significantly higher (no pun intended) than the proportion helped by prochloroperazine. Use a \(1 \%\) significance level since we would require very strong evidence to switch to THC in this case. (b) Why is it important that these data come from a well-designed experiment? $$ \begin{array}{lccc} \hline \text { Treatment } & \text { Sample Size } & \text { Effective } & \text { Not Effective } \\ \hline \text { THC } & 79 & 36 & 43 \\ \text { Prochloroperazine } & 78 & 16 & 62 \\ \hline \end{array} $$

Table 6.29 gives a sample of grades on the first two quizzes in an introductory statistics course. We are interested in testing whether the mean grade on the second quiz is significantly higher than the mean grade on the first quiz. (a) Complete the test if we assume that the grades from the first quiz come from a random sample of 10 students in the course and the grades on the second quiz come from a different separate random sample of 10 students in the course. Clearly state the conclusion. (b) Now conduct the test if we assume that the grades recorded for the first quiz and the second quiz are from the same 10 students in the same order. (So the first student got a 72 on the first quiz and a 78 on the second quiz.) (c) Why are the results so different? Which is a better way to collect the data to answer the question of whether grades are higher on the second quiz? $$ \begin{array}{llllllllll} \hline \text { First Quiz } & 72 & 95 & 56 & 87 & 80 & 98 & 74 & 85 & 77 & 62 \\\ \text { Second Quiz } & 78 & 96 & 72 & 89 & 80 & 95 & 86 & 87 & 82 & 75 \\ \hline \end{array} $$

Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=-2.6, s_{d}=4.1\) \(n_{d}=18\)

The dataset BaseballHits gives 2010 season statistics for all Major League Baseball teams. We treat this as a sample of all MLB teams in all years. Computer output of descriptive statistics for the variable giving the batting average is shown: $$ \begin{aligned} &\text { Descriptive Statistics: BattingAvg }\\\ &\begin{array}{lrrrrr} \text { Variable } & \mathrm{N} & \mathrm{N}^{*} & \text { Mean } & \text { SE Mean } & \text { StDev } \\ \text { BattingAvg } & 30 & 0 & 0.25727 & 0.00190 & 0.01039 \\ \text { Minimum } & & \text { Q1 } & \text { Median } & \text { Q3 } & \text { Maximum } \\ 0.23600 & 0.24800 & 0.25700 & 0.26725 & 0.27600 \end{array} \end{aligned} $$ (a) How many teams are included in the dataset? What is the mean batting average? What is the standard deviation? (b) Use the descriptive statistics above to conduct a hypothesis test to determine whether there is evidence that average team batting average is different from \(0.250 .\) Show all details of the test. (c) Compare the test statistic and p-value you found in part (b) to the computer output below for the same data:

In Exercises 6.159 and \(6.160,\) situations comparing two proportions are described. In each case, determine whether the situation involves comparing proportions for two groups or comparing two proportions from the same group. State whether the methods of this section apply to the difference in proportions. (a) Compare the proportion of students who use a Windows-based \(\mathrm{PC}\) to the proportion who use a Mac. (b) Compare the proportion of students who study abroad between those attending public universities and those at private universities. (c) Compare the proportion of in-state students at a university to the proportion from outside the state. (d) Compare the proportion of in-state students who get financial aid to the proportion of outof-state students who get financial aid.
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