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Chapter 9: Problem 3

What three assumptions must be met when you are using the \(z\) test to test differences between two means when \(\sigma_{1}\) and \(\sigma_{2}\) are known?

Short Answer

Expert verified
The assumptions are random sampling, normality of the distribution of differences, and known population standard deviations.

Step by step solution

01

Random Sampling

The first assumption when using the \(z\) test to compare two means is that the samples must be randomly selected. This ensures each member of the population has an equal chance of being included in the sample, which helps to generalize the results to the entire population.
02

Normality

The second assumption is that the distribution of the difference between the two means (\(\bar{x_1} - \bar{x_2}\)) should be approximately normal. This can be achieved if each sample is large enough (typically \(n \geq 30\)) or if the original populations are normally distributed.
03

Known Population Standard Deviations

The third assumption is that the standard deviations of both populations (\(\sigma_1\) and \(\sigma_2\)) must be known. If they are unknown, the \(z\) test is not appropriate, and a \(t\) test might be more suitable.

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Key Concepts

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

Random Sampling
When conducting a statistical analysis using the \(z\) test, one of the fundamental assumptions is random sampling. Random sampling ensures that each individual or data point in the entire population has an equal probability of being selected for the sample. This equality in selection probability is crucial to achieve unbiased results.

Imagine you're picking marbles from a big jar. If you close your eyes and mix them well before picking, you ensure a fair chance for any marble to be chosen. This is similar to how random sampling works in statistics.

  • Increases the likelihood that the sample accurately represents the population.
  • Helps make results generalizable to the broader group you're interested in.
Failure to use a random sample could lead to skewed results and incorrect conclusions. It鈥檚 important because the integrity of your conclusions hinges on the accuracy of the sampling.
Normality Assumption
A critical aspect of using a \(z\) test is meeting the normality assumption for testing differences between means. The normality assumption implies that the distribution of the difference between the sample means should be approximately normal. This condition is necessary to apply the properties of the normal distribution to your test.

There are a couple of pathways to ensure normality:
  • If each sample's size is large enough (typically \(n \geq 30\)), the Central Limit Theorem suggests that the sampling distribution of the difference in means will be approximately normal.
  • If the original populations themselves are normally distributed, even small samples should reflect normality in their sampling distributions.
Meeting this assumption helps statisticians work with the predictability of the normal curve, making it easier to make valid statistical inferences.
Known Standard Deviations
The requirement for known standard deviations in the \(z\) test is what often distinguishes it from the \(t\) test. For a \(z\) test, it is assumed that the standard deviations of the two populations you鈥檙e examining, \(\sigma_{1}\) and \(\sigma_{2}\), are known.

  • Having known standard deviations helps in calculating the standard error of the means precisely.
  • If the standard deviations are unknown, using the \(z\) test becomes inappropriate.
In cases where standard deviations cannot be precisely obtained, the \(t\) test becomes a more fitting option because it accommodates such scenarios by estimating these values directly from sample data. Ensuring you have accurate measures of variability in your populations enhances the reliability of your testing outcomes.

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Most popular questions from this chapter

Ages of Homes Whiting, Indiana, leads the "Top 100 Cities with the Oldest Houses" list with the average age of houses being 66.4 years. Farther down the list resides Franklin, Pennsylvania, with an average house age of 59.4 years. Researchers selected a random sample of 20 houses in each city and obtained the following statistics. At \(\alpha=0.05,\) can it be concluded that the houses in Whiting are older? Use the \(P\) -value method. $$ \begin{array}{ccc}{} & {\text { Whiting }} & {\text { Franklin }} \\ \hline \text { Mean age } & {62.1 \text { years }} & {55.6 \text { years }} \\\ {\text { Standard deviation }} & {5.4 \text { years }} & {3.9 \text { years }}\end{array} $$

Exam Scores at Private and Public Schools A researcher claims that students in a private school have exam scores that are at most 8 points higher than those of students in public schools. Random samples of 60 students from each type of school are selected and given an exam. The results are shown. At \(\alpha=0.05,\) test the claim. $$ \begin{array}{ll}{\text { Private school }} & {\text { Public school }} \\\ \hline \bar{X_{1}=110} & {\bar{X}_{2}=104} \\ {\sigma_{1}=15} & {\sigma_{2}=15} \\ {n_{1}=60} & {n_{2}=60}\end{array} $$

Find the proportions \(\hat{p}\) and \(\hat{q}\) for each. a. \(n=52, X=32\) b. \(n=80, X=66\) c. \(n=36, X=12\) d. \(n=42, X=7\) e. \(n=160, X=50\)

Gasoline Prices A random sample of monthly gasoline prices was taken from 2011 and from 2015. The samples are shown. Using \(\alpha=0.01,\) can it be concluded that gasoline cost more in 2015? Use the \(P\) -value method. $$ \frac{2011}{2015} | \begin{array}{cccccc}{2011} & {2.02} & {2.47} & {2.50} & {2.70} & {3.13} & {2.56} \\ \hline 2015 & {2.36} & {2.46} & {2.63} & {2.76} & {3.00} & {2.85} & {2.77}\end{array} $$

Sale Prices for Houses The average sales price of new one-family houses in the Midwest is dollar 250,000 and in the South is dollar 253,400. A random sample of 40 houses in each region was examined with the following results. At the 0.05 level of significance, can it be concluded that the difference in mean sales price for the two regions is greater than dollar 3400 ? $$ \begin{array}{lcc}{} & {\text { South }} & {\text { Midwest }} \\ \hline \text { Sample size } & {40} & {40} \\ {\text { Sample mean }} & {\$ 261,500} & {\$ 248,200} \\ {\text { Population standard deviation }} & {\$ 10,500} & {\$ 12,000}\end{array} $$
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