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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 three assumptions are normality of distribution, known population variances, and independent samples.

Step by step solution

01

Understanding the Assumptions

Before using the \(z\) test to test the difference between two means, it is crucial to be aware of the necessary assumptions. These assumptions ensure the validity of the results derived from the \(z\) test. Each assumption addresses a particular aspect of the data distribution or sampling process.
02

Assumption 1 - Normality of Distribution

The first assumption is that the data from each group should be drawn from a normally distributed population. This means that the distribution of the data around the mean for both groups should follow the normal distribution (bell-shaped curve), especially when dealing with small sample sizes. However, with a large sample size, the Central Limit Theorem implies normality.
03

Assumption 2 - Known Population Variances

The second assumption is that the population variances, \(\sigma_1^2\) and \(\sigma_2^2\), are known. This is essential for calculating the standard error of the difference in means accurately. When variances are unknown and need estimation from the sample, a different test, such as the \(t\) test, may be more appropriate.
04

Assumption 3 - Independent Samples

The third assumption is that the samples from the two populations are independent. This means that the data collected from one group should not influence or be related to the data from the other group. Independence ensures that the sample values are unbiased and the \(z\) test can validly assess the difference between groups.

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Key Concepts

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

Normal Distribution
When using a \(z\) test, one pivotal assumption is that the data collected comes from a normally distributed population. In simpler terms, this means that if you were to plot a graph based on the data, it would resemble a bell-shaped curve. This shape is indicative of a normal distribution. Why does this matter? Well, in statistics, the normal distribution provides a foundation for making predictions about data trends.

Here are some key points about normal distribution:
  • Data points tend to be symmetrically distributed around the mean.
  • Most values lie close to the mean, with fewer situated at either extreme end of the scale.
In practice, even if the population itself isn't perfectly normal, the Central Limit Theorem comes to the rescue. It indicates that for large sample sizes, the distribution of the sample mean will resemble a normal distribution, even if the population distribution is not perfectly normal. This is especially reassuring when working with diverse data sets.
Population Variances
Another cornerstone of the \(z\) test is the necessity to know the population variances, often represented as \(\sigma_1^2\) and \(\sigma_2^2\). Knowing these variances allows for an accurate calculation of the standard error of the difference between two means.

But what exactly are population variances?
  • They measure how much the data in each population varies from the mean.
  • Smaller variances indicate that the data points are clustered closely around the mean, while larger variances imply more spread out data.
Why is this assumption crucial for a \(z\) test? Without knowing population variances, the reliability of the results may be compromised. When variances are unknown, it is advisable to consider alternative methods like the \(t\) test, which estimates variance from the sample data.
Independent Samples
The assumption of independent samples is a vital requirement for the \(z\) test. It states that the samples drawn from two different populations must be independent of one another. Simply put, the outcome or results from one sample should not affect or be related to the results from the other sample.

Here's why it matters:
  • Independent samples ensure unbiased results, permitting valid comparative analysis.
  • If samples are dependent, any inference made about the population parameters may be flawed.
This concept becomes almost second nature in statistics. You always want to ensure that your comparison between groups stands on solid ground, and independence of samples enables that.

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Most popular questions from this chapter

Adults aged 16 or older were assessed in three types of literacy: prose, document, and quantitative. The scores in document literacy were the same for 19 - to 24 -year-olds and for 40 - to 49 -year-olds. A random sample of scores from a later year showed the following statistics. $$ \begin{array}{lccc} & & \text { Population } & \\ \text { Age group } & \begin{array}{l} \text { Mean } \\ \text { score } \end{array} & \begin{array}{c} \text { standard } \\ \text { deviation } \end{array} & \begin{array}{c} \text { Sample } \\ \text { size } \end{array} \\ \hline 19-24 & 280 & 56.2 & 40 \\ 40-49 & 315 & 52.1 & 35 \end{array} $$ Construct a \(95 \%\) confidence interval for the true difference in mean scores for these two groups. What does your interval say about the claim that there is no difference in mean scores?

What are the two different degrees of freedom associated with the \(F\) distribution?

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A random sample of daily high temperatures in January and February is listed. At \(\alpha=0.05,\) can it be concluded that there is a difference in variances in high temperature between the two months? $$ \begin{array}{l|cccccccccc} \text { Jan. } & 31 & 31 & 38 & 24 & 24 & 42 & 22 & 43 & 35 & 42 \\ \hline \text { Feb. } & 31 & 29 & 24 & 30 & 28 & 24 & 27 & 34 & 27 & \end{array} $$

Perform each of these steps. Assume that all variables are normally or approximately normally distributed a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Mistakes in a Song A random sample of six music students played a short song, and the number of mistakes in music each student made was recorded. After they practiced the song 5 times, the number of mistakes each student made was recorded. The data are shown. At \(\alpha=0.05,\) can it be concluded that there was a decrease in the mean number of mistakes? $$ \begin{array}{l|rrrrrr} \text { Student } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } \\ \hline \text { Before } & 10 & 6 & 8 & 8 & 13 & 8 \\ \hline \text { After } & 4 & 2 & 2 & 7 & 8 & 9 \end{array} $$

Perform each of these steps. Assume that all variables are normally or approximately normally distributed a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Toy Assembly Test An educational researcher devised a wooden toy assembly project to test learning in 6 -year-olds. The time in seconds to assemble the project was noted, and the toy was disassembled out of the child's sight. Then the child was given the task to repeat. The researcher would conclude that learning occurred if the mean of the second assembly times was less than the mean of the first assembly times. At \(\alpha=0.01,\) can it be concluded that learning took place? Use the \(P\) -value method, and find the \(99 \%\) confidence interval of the difference in means $$ \begin{array}{l|rrrrrrr} \text { Child } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text { Trial 1 } & 100 & 150 & 150 & 110 & 130 & 120 & 118 \\ \hline \text { Trial 2 } & 90 & 130 & 150 & 90 & 105 & 110 & 120 \end{array} $$

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