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Chapter 10: Problem 33

Normal assumption The methods of this section make the assumption of a normal population distribution. Why do you think this is more relevant for small samples than for large samples? (Hint: What shape does the sampling distribution of \(\bar{x}_{1}-\bar{x}_{2}\) have for large samples, regardless of the actual shape of the population distributions?)

Short Answer

Expert verified
For small samples, normality is more crucial as the Central Limit Theorem doesn't ensure normality in the sampling distribution. For large samples, normality is less relevant due to the Central Limit Theorem.

Step by step solution

01

Understanding Normal Assumption

The normal assumption implies that the data originates from a normally distributed population. This is important in various statistical methods because many tests, such as the t-test, rely on this assumption for accurate results.
02

Relevance for Small Samples

For small sample sizes, the Central Limit Theorem doesn't strongly apply, meaning the sampling distribution of the sample mean may not approximate normality if the population distribution is not normal. Therefore, it's crucial that the population is normally distributed to make valid inferences.
03

Relevance for Large Samples

For large sample sizes, the Central Limit Theorem states that regardless of the population's distribution, the sampling distribution of the difference between sample means, \(\bar{x}_1 - \bar{x}_2\), will become approximately normally distributed. This makes the normal assumption less critical.
04

Connecting Assumptions to Sample Size

Small samples won't benefit from the Central Limit Theorem to smooth out irregularities in the population distribution, making normality assumptions more critical to ensure accurate statistical testing. However, with larger samples, the law of large numbers helps ensure a normal distribution of sample means, thus reducing the reliance on the initial assumption of normality.

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Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in statistics. Imagine a bell-shaped curve that shows how data points are spread around the mean (average).
Normal distribution is symmetrical, which means the left side of the curve is a mirror image of the right side.
  • Many natural phenomena tend to follow a normal distribution, such as heights, test scores, and more.
  • It's essential for statistical methods because it allows for ease of calculation and inference.
The key characteristic of a normal distribution is that it provides a good model for inferential statistics.
This is because in a normal distribution most of the data fall around the mean, making predictions and testing more valid. This is why statistical tests often assume normality: it simplifies analyzing the data.
Sampling Distribution
A sampling distribution shows how a statistic, such as a sample mean, is distributed over numerous samples taken from the same population.
It's a critical concept that bridges the population and the sample.
  • When we repeatedly take samples from a population, the distribution of a statistic from those samples forms its own distribution.
  • The Central Limit Theorem tells us that this sampling distribution becomes more normal as the sample size increases.
In many studies, researchers use a sample from the population and analyze it rather than studying the entire population.
This distribution helps to highlight the variability of sample statistics and informs how closely we can expect a sample mean to match the actual population mean.
Understanding it is key to making correct inferences from the sample to the population.
T-Test
A t-test is a statistical test used to compare the means of two groups.
It helps to determine if the observed difference is statistically significant or if it could have occurred by chance.
  • There are different types of t-tests, such as independent t-tests for comparing two different groups and dependent t-tests for comparing the same group at different times.
  • The t-test assumes that the data are normally distributed, which is why normal distribution is important.
The t-test is widely used because it allows conclusions to be drawn about the population from which the samples originate.
This is especially crucial when sample sizes are small, as it ensures reliable and valid results under the framework of assumed normality.
For large samples, the Central Limit Theorem helps relax the normality condition.
Sample Size
Sample size refers to the number of observations in a sample.
It's a vital aspect of statistical testing because it affects the precision and accuracy of the sample results.
  • A larger sample size generally provides a more accurate reflection of the population, reducing error and increasing confidence in the findings.
  • Small sample sizes might not capture the true variability of the population, potentially leading to incorrect conclusions.
In smaller samples, deviations from a normal distribution are more pronounced due to less averaging effect, placing more importance on normality assumptions.
Conversely, with larger samples, the Central Limit Theorem allows the sampling distribution to approximate normalcy, reducing reliance on the population鈥檚 initial distribution.
This makes sample size a critical consideration when planning and interpreting statistical analyses.

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Most popular questions from this chapter

The following data refer to a random sample of prize money earned by male and female skiers racing in the \(2014 / 2015\) FIS world cup season (in Swiss Franc). Males: \(\quad 89000,179000,8820,12000,10750,66000,\) 6700,3300,74000,56800 Females: \(\quad 73000,95000,32400,4000,2000,57100,4500\) Enter the observations (separated by spaces) into the Permutation Test web app. Let male skiers be Group 1 and female skiers be Group 2 and let the test statistic be the difference in sample means. Is there evidence that male skiers earn more, on average, than female skiers? a. What are the observed group means and their difference? (The subtitle of the dot plot shows this information.) b. Why would we prefer to run a permutation test over a \(t\) test? c. Press the Generate Random Permutation(s) button once to generate one permutation of the original data. What are the two group means and their difference under this permutation? d. Did this one permutation lead to a difference that is less extreme or more extreme than the observed difference? e. Click Generate Random Permutation(s) nine more times, for a total of 10 permutations. How many of them resulted in a test statistic at least as extreme as the observed one? f. Select to generate 10,000 random permutations. How many of them resulted in a test statistic as or more extreme? g. Find and interpret the permutation P-value. h. Select the option to generate all possible permutations. Do you notice a big difference in the histogram and P-value based on 10,000 randomly sampled permutations and on all possible permutations?

Two TV commercials are developed for marketing a new product. A volunteer test sample of 200 people is randomly split into two groups of 100 each. In a controlled setting, Group A watches commercial A and Group B watches commercial B. In Group A, 25 say they would buy the product. In group \(\mathrm{B}, 20\) say they would buy the product. The marketing manager who devised this experiment concludes that commercial \(\mathrm{A}\) is better. Is this conclusion justified? Analyze the data. If you prefer, use software (such as MINITAB or a web app) for which you can enter summary counts. a. Show all steps of your analysis (perhaps including an appropriate graph) and check assumptions. b. Comment on the manager's conclusion and indicate limitations of the experiment.

An AP story (April 9,2005\()\) with headline Study: Attractive People Make More stated that "A study concerning weight showed that women who were obese earned 17 percent lower wages than women of average weight." a. Identify the two variables stated to have an association. b. Identify a control variable that might explain part or all of this association. If you had the original data including data on that control variable, how could you check whether the control variable does explain the association?

A study of a sample of horseshoe crabs on a Florida island (J. Brockmann, Ethology, vol. \(102,1996,\) pp. \(1-21\) ) investigated the factors that were associated with whether female crabs had a male crab mate. Basic statistics, including the five-number summary on weight (kg) for the 111 female crabs who had a male crab nearby and for the 62 female crabs who did not have a male crab nearby, are given in the table. Assume that these horseshoe crabs have the properties of a random sample of all such crabs. $$\begin{array}{lcccccccc}\hline {\text { Summary Statistics for Weights of Horseshoe Crabs }} \\ \hline & n & \text { Mean } & \text { Std. Dev. } & \text { Min } & \text { Q1 } & \text { Med } & \text { Q3 } & \text { Max } \\\\\hline \text { Mate } & 111 & 2.6 & 0.6 & 1.5 & 2.2 & 2.6 & 3.0 & 5.2 \\ \text { No Mate } & 62 & 2.1 & 0.4 & 1.2 & 1.8 & 2.1 & 2.4 & 3.2 \\ \hline\end{array}$$ a. Sketch box plots for the weight distributions of the two groups. Interpret by comparing the groups with respect to shape, center, and variability. b. Estimate the difference between the mean weights of female crabs who have mates and female crabs who do not have mates. c. Find the standard error for the estimate in part b. d. Construct a \(90 \%\) confidence interval for the difference between the population mean weights, and interpret.

In \(2015,\) a survey of firstyear university students in Brazil was conducted to determine if they knew how to activate the Mobile Emergency Attendance Service (MEAS). Of the 1038 respondents (59.5\% studying biological sciences, \(11.6 \%\) physical sciences, and \(28.6 \%\) humanities) \(, 54.3 \%\) students of nonbiological subjects \((n=564)\) knew how to activate the MEAS as compared to \(61.4 \%\) students of biological sciences \((n=637)\). (Source: https://www.ncbi.nlm.nih.gov/ \(\mathrm{pmc/articles} / \mathrm{PMC} 4661033 /)\) a. Estimate the difference between the proportions of students of biological sciences and nonbiological subjects who know how to activate the MEAS and interpret. b. Find the standard error for this difference. Interpret it. c. Define the two relevant population parameters for comparison in the context of this exercise. d. Construct and interpret a \(95 \%\) confidence interval for the difference in proportions, explaining how your interpretation reflects whether the interval contains \(0 .\) e. State and check the assumptions for the confidence interval in part d to be valid.
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