91Ó°ÊÓ

Annual per capita consumption of milk is 21.6 gallons (Statistical Abstract of the United States: 2006 ). Being from the Midwest, you believe milk consumption is higher there and wish to support your opinion. A sample of 16 individuals from the midwestern town of Webster City showed a sample mean annual consumption of 24.1 gallons with a standard deviation of \(s=4.8\) a. Develop a hypothesis test that can be used to determine whether the mean annual consumption in Webster City is higher than the national mean. b. What is a point estimate of the difference between mean annual consumption in Webster City and the national mean? c. \(\quad\) At \(\alpha=.05,\) test for a significant difference. What is your conclusion?

Step by step solution

01

Define the Hypotheses

We need to determine whether the mean annual milk consumption in Webster City is higher than the national mean. We set up the hypotheses as follows: \(H_0: \mu = 21.6\) (the mean consumption is the same as the national mean) and \(H_a: \mu > 21.6\) (the mean consumption in Webster City is greater).

02

Identify the Test Statistic

For this problem, we will use a t-test for the sample mean. The test statistic is calculated using the formula \( t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \), where \( \bar{x} = 24.1 \), \( \mu_0 = 21.6 \), \( s = 4.8 \), and \( n = 16 \).

03

Calculate the Test Statistic

Substitute the given values into the formula: \( t = \frac{24.1 - 21.6}{4.8/\sqrt{16}} = \frac{2.5}{1.2} = 2.083 \).

04

Determine the Critical Value and Rejection Region

We use a one-tailed t-test with \( \alpha = 0.05 \) and degrees of freedom \( df = n - 1 = 15 \). From the t-distribution table, the critical value \( t_{0.05, 15} \approx 1.753 \). The rejection region is \( t > 1.753 \).

05

Make a Decision

Since the calculated test statistic \( t = 2.083 \) is greater than the critical value \( 1.753 \), we reject the null hypothesis \( H_0 \). Thus, there is significant evidence at the \( \alpha = 0.05 \) level to conclude that the mean annual consumption in Webster City is higher than the national mean.

06

Calculate the Point Estimate of the Difference

The point estimate of the difference between the mean annual consumption in Webster City and the national mean is \( \bar{x} - \mu_0 = 24.1 - 21.6 = 2.5 \) gallons.

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Key Concepts

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

t-test

The t-test is a statistical method that helps us determine if there is a significant difference between the means of two groups. In simple terms, it helps us understand if the difference we see in data is due to chance or a genuine variation.
In this exercise, we are interested in checking if the mean milk consumption in Webster City is higher than the national average.
We'll use a one-sample t-test because we're comparing the sample mean from Webster City with the known national mean. The one-sample t-test uses the formula:

• Solution Formula: \( t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \)

Where:

• \( \bar{x} \) is the sample mean,
• \( \mu_0 \) is the national mean,
• \( s \) is the standard deviation of the sample,
• \( n \) is the sample size.

By applying this formula, we can compute the t-value and understand whether the observed difference is statistically significant.

test statistic

The test statistic is a number that is computed from the sample data. It allows us to make a decision about the null hypothesis in hypothesis testing.
In simpler terms, the test statistic helps determine how far our data is from what we expect under the null hypothesis.
In our example, the test statistic is calculated using the formula mentioned in the t-test section. When we plug in the numbers:

• Sample mean \( \bar{x} = 24.1 \)
• National mean \( \mu_0 = 21.6 \)
• Standard deviation \( s = 4.8 \)
• Sample size \( n = 16 \)

We get:

• \( t = \frac{24.1 - 21.6}{4.8/\sqrt{16}} = 2.083 \)

This value helps us decide if the difference between Webster City's milk consumption and the national average is significant enough to reject the null hypothesis.

critical value

The critical value in hypothesis testing is a threshold that our test statistic must exceed to reject the null hypothesis. It's derived from the t-distribution or z-distribution depending on the test type and size.
In our case, we refer to a t-distribution table because we're conducting a t-test, and we look for the critical value based on:

• The chosen significance level \( \alpha = 0.05 \)
• The degrees of freedom, which is calculated as \( df = n - 1 = 15 \)

For a one-tailed test with these parameters, the critical value is approximately 1.753.
To make a decision, we compare the test statistic with this critical value. Because our test statistic \( t = 2.083 \) is greater than the critical value \( 1.753 \), we reject the null hypothesis, supporting the claim that milk consumption in Webster City is higher than the national average.

point estimate

In statistics, a point estimate is a single value calculated from sample data used to estimate a population parameter. It's basically our best guess of what the true average difference might be.
In this exercise, we're interested in the difference between the mean annual milk consumption in Webster City and the national mean.
The solution calculates the point estimate as:

• \( \bar{x} - \mu_0 = 24.1 - 21.6 = 2.5 \)

So, the point estimate of the difference is 2.5 gallons.
This tells us that, on average, people in Webster City drink 2.5 gallons more milk annually than the national average.
A point estimate gives us a clear, succinct measure of difference, though its accuracy can be further corroborated with a confidence interval.