91Ó°ÊÓ

Two populations have normal distributions. The first has population standard deviation 2 and the second has population standard deviation \(3 .\) A random sample of 16 measurements from the first population had a sample mean of \(20 .\) An independent random sample of 9 measurements from the second population had a sample mean of \(19 .\) Test the claim that the population mean of the first population exceeds that of the second. Use a \(5 \%\) level of significance. (a) Check Requirements What distribution does the sample test statistic follow? Explain. (b) State the hypotheses. (c) Compute \(\bar{x}_{1}-\bar{x}_{2}\) and the corresponding sample distribution value. (d) Find the \(P\) -value of the sample test statistic. (e) Conclude the test (f) Interpret the results.

Step by step solution

01

Check Requirements

Since both populations have normal distributions and we know the population standard deviations, we will use a normal distribution to assess the test statistic for the difference in sample means. The appropriate test is a two-sample z-test.

02

State the Hypotheses

The null hypothesis ( H_0 ) is that the mean of the first population is equal to or less than that of the second population ( ar{ u}_1 = ar{ u}_2 or ≤ ar{ u}_2 ). The alternative hypothesis ( H_1 ) is that the mean of the first population exceeds that of the second population ( ar{ u}_1 > ar{ u}_2 ).

03

Compute Sample Mean Difference and Test Statistic

Compute the difference in sample means: ar{x}_1 - ar{x}_2 = 20 - 19 = 1. The standard deviation for the difference in sample means is \( \sqrt{\frac{{2^2}}{16} + \frac{{3^2}}{9}} = \sqrt{0.25+1} = \sqrt{1.25} = 1.118 \). Thus, the z-test statistic is \( z = \frac{{1}}{{1.118}} \approx 0.895 \).

04

Find the P-value

To find the P-value, we look at the standard normal distribution table. For z = 0.895 , the P-value is approximately 0.185 (considering it's a one-tailed test).

05

Conclude the Test

The P-value of 0.185 exceeds the significance level of 0.05 , which means we fail to reject the null hypothesis. This implies there isn't sufficient evidence to support the claim that the first population's mean is greater.

06

Interpret the Results

At a 5% level of significance, the data does not provide enough evidence to claim that the mean of the first population exceeds the mean of the second population.

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Key Concepts

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

z-test

A z-test is a statistical method used to determine if there is a significant difference between the means of two populations. It is particularly useful when the population standard deviations are known, as they are in this case. The z-test helps us decide if the observed difference between sample means reflects a true difference in the population means.

• **Two-sample z-test:** This is the test we employed in this problem since it compares the difference between two independent sample means.
• **Z-test formula:** The z-score is calculated by dividing the difference in sample means by the standard error of the difference between the means.

Understanding how to compute the z-score is essential. It gives us a measure of how many standard deviations the observed difference is from the expected difference under the null hypothesis. A high z-score implies a large difference, leading potentially to rejecting the null hypothesis.

normal distribution

The concept of normal distribution is fundamental to hypothesis testing and statistics. It describes a distribution that is symmetric, with most observations centering around the mean and tapering off as they move away.

• **Characteristics:** It is described by its bell-shaped curve, which is defined by its mean and standard deviation.
• **Importance in testing:** Normal distributions allow us to predict probabilities and are pivotal when using z-tests, as the test statistic follows this distribution.

In the context of our exercise, both populations are normal, meaning the conditions for using a z-test are met. The test can proceed as we assume the sampling distribution of the sample mean difference follows a normal distribution thanks to the Central Limit Theorem. This theorem states that sample means will tend to have a normal distribution, especially as sample sizes increase.

p-value

The p-value in hypothesis testing is a crucial concept that informs us about the significance of our results. It helps determine the strength of evidence against the null hypothesis.

• **Definition:** The p-value is the probability of obtaining a test statistic equal to or more extreme than the observed value, assuming the null hypothesis is true.
• **Interpretation:** A low p-value (typically ≤ 0.05) suggests significant evidence against the null hypothesis, while a higher p-value indicates weak evidence against it.

For our problem, a p-value of 0.185 was calculated, which is greater than 0.05. This tells us that the evidence isn't strong enough to reject the null hypothesis, meaning the difference in means isn't statistically significant at the chosen level of significance.

level of significance

The level of significance, often denoted by α, is a threshold set by the researcher to determine whether the test statistic is extreme enough to reject the null hypothesis. It is a predetermined probability of committing a Type I error, which involves incorrectly rejecting the null hypothesis.

• **Common choices:** Typical levels of significance are 0.05, 0.01, and 0.10. A 5% level, used in our exercise, signifies a 5% risk of finding a difference when there is none.
• **Decision making:** If the p-value falls below this level, we have grounds to reject the null hypothesis. However, if it is above the level, as in our scenario, we fail to reject the null hypothesis.

Setting the level of significance safeguards us against making decisions based on random chance rather than actual effects or differences. In hypothesis testing, understanding this balance is key to interpreting results logically and accurately.