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Chapter 6: Problem 250

Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) using the sample results \(\bar{x}_{1}=56, s_{1}=8.2\) with \(n_{1}=30\) and \(\bar{x}_{2}=51, s_{2}=6.9\) with \(n_{2}=40\)

Short Answer

Expert verified
The result of the hypothesis test will depend on the calculated values and the comparison of the test statistic to the critical value. If the test statistic is greater than the critical value, the null hypothesis is rejected in favor of the alternative hypothesis; otherwise, the null hypothesis cannot be rejected.

Step by step solution

01

Calculate the Pooled Standard Deviation

The formula for pooled standard deviation (s_p) is given by: \[s_p=\sqrt{\frac{(n_{1}-1)s_{1}^2+(n_{2}-1)s_{2}^2}{n_{1}+n_{2}-2}}\] Here, \(n_{1}=30, s_{1}=8.2, n_{2}=40, s_{2}=6.9\). Substituting these values gives the pooled standard deviation.
02

Find the Test Statistic

The formula for the test statistic (t) is given by: \[t=\frac{\bar{x}_{1} - \bar{x}_{2}}{s_p \cdot \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}\] Here, \(\bar{x}_{1}=56, \bar{x}_{2}=51\) and use the pooled standard deviation from Step 1.
03

Determine the Degrees of Freedom

In a two-sample t-test, the degrees of freedom is given by: \[df= n_{1}+n_{2}-2\] Here, substituting \(n_{1}=30, n_{2}=40\) gives the degrees of freedom.
04

Find the Critical Value

Use the t-distribution table to find the critical value for the given degrees of freedom and the significance level (usually 0.05 for a two-tailed test). Since this is a one-tailed test, we use 0.05 directly.
05

Test the Hypothesis

Compare the absolute value of the test statistic with the critical value. If the test statistic is larger than the critical value, we reject the null hypothesis in favor of the alternative hypothesis. Otherwise, we fail to reject the null hypothesis.

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Key Concepts

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

Pooled Standard Deviation
When conducting a two-sample t-test, the pooled standard deviation is a crucial component.
It serves as a combined measure of the spread or variability of the two sample groups.
This value helps in determining the test statistic, which ultimately assists in hypothesis testing.
To calculate it, we use the formula:\[s_p=\sqrt{\frac{(n_{1}-1)s_{1}^2+(n_{2}-1)s_{2}^2}{n_{1}+n_{2}-2}}\]where:
  • \(s_1\) and \(s_2\) are the standard deviations of samples 1 and 2 respectively,
  • \(n_1\) and \(n_2\) are the sample sizes.
The pooled standard deviation accounts for different sample sizes by weighing each group's variance by their respective degrees of freedom.
In simpler terms, it averages out the variation between two samples, thus providing a common baseline for comparison.
Understanding and correctly calculating the pooled standard deviation is vital for precise hypothesis testing results.
Hypothesis Testing
Hypothesis testing is a method used to decide whether to reject or fail to reject the null hypothesis.
It involves comparing a test statistic calculated from sample data to a critical value derived from a probability distribution, typically the t-distribution for small sample sizes.
In the given exercise, the hypotheses are:
  • Null Hypothesis (\(H_0\)): \(\mu_1 = \mu_2\), meaning there is no difference between the mean of the two groups.
  • Alternative Hypothesis (\(H_a\)): \(\mu_1 > \mu_2\), suggesting that the mean of group 1 is greater than group 2.
The process includes:
  • Calculating the test statistic using both sample means and the pooled standard deviation.
  • Determining the critical value from the t-distribution table for a given confidence level.
  • Comparing the test statistic to this critical value to make a decision on the null hypothesis.
If the test statistic exceeds the critical value in an appropriate direction, we reject the null hypothesis, indicating a statistically significant difference between the groups.
Otherwise, we conclude that there is not enough evidence to support the alternative hypothesis.
Degrees of Freedom
Degrees of freedom (df) refer to the number of independent values that are free to vary in the analysis.
They play a significant role in determining the critical value from the t-distribution in hypothesis testing.
For a two-sample t-test, the degrees of freedom can be calculated as follows:\[df = n_1 + n_2 - 2\]where:
  • \(n_1\) is the sample size of the first group,
  • \(n_2\) is the sample size of the second group.
In this exercise, replacing \(n_1\) with 30 and \(n_2\) with 40 provides the degrees of freedom to be 68.
Understanding degrees of freedom is essential as it influences the shape of the t-distribution curve, which affects the critical value.
A larger degree of freedom generally leads to a narrower distribution, resulting in a more precise critical value.

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Most popular questions from this chapter

Data 1.3 on page 10 discusses a study designed to test whether applying a metal tag is detrimental to a penguin, as opposed to applying an electronic tag. One variable examined is the date penguins arrive at the breeding site, with later arrivals hurting breeding success. Arrival date is measured as the number of days after November 1st. Mean arrival date for the 167 times metal- tagged penguins arrived was December 7 th ( 37 days after November 1 st ) with a standard deviation of 38.77 days, while mean arrival date for the 189 times electronic-tagged penguins arrived at the breeding site was November 21 st (21 days after November 1 st ) with a standard deviation of \(27.50 .\) Do these data provide evidence that metal-tagged penguins have a later mean arrival time? Show all details of the test.

Use a t-distribution. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the endpoints of the t-distribution with \(2.5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=15\) and \(n_{2}=25\)

A sample with \(n=10, \bar{x}=508.5,\) and \(s=21.5\)

Green Tea and Prostate Cancer A preliminary study suggests a benefit from green tea for those at risk of prostate cancer. \(^{55}\) The study involved 60 men with PIN lesions, some of which turn into prostate cancer. Half the men, randomly determined, were given \(600 \mathrm{mg}\) a day of a green tea extract while the other half were given a placebo. The study was double-blind, and the results after one year are shown in Table \(6.14 .\) Does the sample provide evidence that taking green tea extract reduces the risk of developing prostate cancer? $$ \begin{array}{lcc} \hline \text { Treatment } & \text { Cancer } & \text { No Cancer } \\ \hline \text { Green tea } & 1 & 29 \\ \text { Placebo } & 9 & 21 \end{array} $$

The dataset ICUAdmissions contains information on a sample of 200 patients being admitted to the Intensive Care Unit (ICU) at a hospital. One of the variables is HeartRate and another is Status which indicates whether the patient lived (Status \(=0\) ) or died (Status \(=1\) ). Use the computer output to give the details of a test to determine whether mean heart rate is different between patients who lived and died. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) $$ \begin{aligned} &\text { Two-sample } \mathrm{T} \text { for HeartRate }\\\ &\begin{array}{lrrrr} \text { Status } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 0 & 160 & 98.5 & 27.0 & 2.1 \\ 1 & 40 & 100.6 & 26.5 & 4.2 \end{array} \end{aligned} $$ Difference \(=m u(0)-m u(1)\) Estimate for difference: -2.12 \(95 \% \mathrm{Cl}\) for difference: (-11.53,7.28) T-Test of difference \(=0(\) vs not \(=):\) T-Value \(=-0.45\) P-Value \(=0.653 \quad \mathrm{DF}=60\)
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