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Chapter 6: Problem 274

Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=15.7, s_{d}=\) 12.2 \(, n_{d}=25\)

Short Answer

Expert verified
The result of this hypothesis test depends on the calculated t-statistic and the corresponding P-value. If the P-value is less than the significance level (usually 0.05), then we reject the null hypothesis; otherwise, we fail to reject it.

Step by step solution

01

Calculate the Standard Error

The standard error (SE) is calculated as the standard deviation of the differences divided by the square root of the number of pairs. Here it is \(SE = \frac{s_d}{\sqrt{n_d}} = \frac{12.2}{\sqrt{25}} \).
02

Calculate the t-statistic

The t-statistic is calculated as the sample mean difference divided by the standard error. Here it is \(t = \frac{\bar{x}_d}{SE}\).
03

Determine the Degree of Freedom

Degree of freedom is calculated as the total number of pairs minus 1. Here it is \(df = n_d - 1 = 25 - 1 = 24\).
04

Determine the t-critical value and P-value

Depending on the calculated t-statistic and the degree of freedom, look up in the t-distribution table to find the critical t-value and P-value. This part allows us to make a decision whether to reject or 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.

Standard Error
The standard error (SE) measures the average distance that sample means differ from the actual population mean. It helps determine the precision of the sample mean as an estimate of the population mean.

In the context of a paired difference test, the standard error tells us how much we can expect the differences in paired observations to vary. It's calculated with the formula:

\[ SE = \frac{s_d}{\sqrt{n_d}} \]

where:
  • \( s_d \) is the standard deviation of the differences,
  • \( n_d \) is the number of pairs.
A lower SE indicates a more precise estimate of the mean difference. In practical terms, this means smaller variability in the pairwise differences, which translates to greater confidence in the sample results.
t-statistic
The t-statistic is a crucial part of hypothesis testing. It allows us to determine whether the observed data is significantly different from what we would expect under the null hypothesis. For a paired sample test, it tells us how many standard errors the sample mean difference is away from the null hypothesis mean difference (usually zero).

The t-statistic is calculated using the formula:

\[ t = \frac{\bar{x}_d}{SE} \]

where:
  • \( \bar{x}_d \) is the sample mean difference,
  • \( SE \) is the standard error.
A larger absolute value of the t-statistic indicates that the sample mean difference is far from the hypothesized mean difference. In hypothesis testing, we compare this calculated t-value against a critical t-value from the t-distribution table to decide whether our sample provides enough evidence to reject the null hypothesis.
Degrees of Freedom
Degrees of freedom (df) play a central role in statistical testing, especially with the t-distribution. They refer to the number of values in a calculation that have the freedom to vary. In simple terms, it's the number of independent pieces of information available after accounting for constraints or fixed parameters.

For the paired difference test, the degrees of freedom are calculated as one less than the number of paired observations:

\[ df = n_d - 1 \]

where \( n_d \) is the number of pairs. This measure ensures the t-distribution accurately reflects the sample size used in the test.

Understanding degrees of freedom is essential because it influences the shape and critical values of the t-distribution. Knowing how to calculate and use degrees of freedom allows us to make informed decisions about hypothesis testing results based on the sample data.

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Most popular questions from this chapter

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:

Number of Bedrooms in Houses in New York and New Jersey The dataset HomesForSale has data on houses available for sale in three Mid-Atlantic states (NY, NJ, and PA). For this exercise we are specifically interested in homes for sale in New York and New Jersey. We have information on 30 homes from each state and observe the proportion of homes with more than three bedrooms. We find that \(26.7 \%\) of homes in NY \(\left(\hat{p}_{N Y}\right)\) and \(63.3 \%\) of homes in NJ \(\left(\hat{p}_{N J}\right)\) have more then three bedrooms. (a) Is the normal distribution appropriate to model this difference? (b) Test for a difference in proportion of homes with more than three bedrooms between the two states and interpret the result.

Exercise 4.86 on page 263 introduces a matched pairs study in which 47 participants had cell phones put on their ears and then had their brain glucose metabolism (a measure of brain activity) measured under two conditions: with one cell phone turned on for 50 minutes (the "on" condition) and with both cell phones off (the "off" condition). Brain glucose metabolism is measured in \(\mu \mathrm{mol} / 100 \mathrm{~g}\) per minute, and the differences of the metabolism rate in the on condition minus the metabolism rate in the off condition were computed for all participants. The mean of the differences was 2.4 with a standard deviation of \(6.3 .\) Find and interpret a \(95 \%\) confidence interval for the effect size of the cell phone waves on mean brain metabolism rate.

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? $$ \begin{array}{lcc} \hline \text { Treatment } & \text { Cancer } & \text { No Cancer } \\ \hline \text { Green tea } & 1 & 29 \\ \text { Placebo } & 9 & 21 \end{array} $$

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.
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