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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?

Short Answer

Expert verified
Milk consumption in Webster City is significantly higher than the national average at \(\alpha = 0.05\).

Step by step solution

01

Define Hypotheses

To test the claim that milk consumption in Webster City is higher than the national average of 21.6 gallons, we perform a hypothesis test. The null hypothesis (\(H_0\)) is that the mean annual milk consumption in Webster City is equal to the national mean, i.e., \(H_0: \mu = 21.6\). The alternative hypothesis (\(H_a\)) is that the mean annual milk consumption in Webster City is greater than the national mean, i.e., \(H_a: \mu > 21.6\).
02

Point Estimate of Difference

The point estimate of the difference between the mean annual consumption in Webster City and the national mean is the sample mean minus the national mean. This is calculated as \(24.1 - 21.6 = 2.5\) gallons.
03

Determine Test Statistic

We use the t-distribution to conduct this test because the population standard deviation is unknown and the sample size is small (less than 30). The test statistic \(t\) 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 \(n = 16\). Substituting these values in gives us: \[ t = \frac{24.1 - 21.6}{4.8/\sqrt{16}} \approx \frac{2.5}{1.2} \approx 2.083 \]
04

Find Critical Value and Decision Rule

With a significance level of \(\alpha = 0.05\) and \(n - 1 = 15\) degrees of freedom, we refer to the t-distribution table to find the critical value \(t_{0.05, 15}\). The critical value is approximately 1.753. The decision rule is: reject \(H_0\) if the calculated \(t\) value is greater than the critical value.
05

Conclusion

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

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Key Concepts

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

t-distribution
When conducting a hypothesis test, especially with small sample sizes, the t-distribution becomes a critical tool. Unlike the normal distribution, which is symmetrical and applies when population parameters are known or sample sizes are large, the t-distribution accounts for variability with unknown population standard deviations.
If you don't know the population standard deviation but need to perform a hypothesis test, and your sample size is less than 30, the t-distribution is your friend. It helps to compensate for the increased variability and gives us a more reliable test statistic.
  • The t-distribution is also symmetrical but has heavier tails, meaning it is more prone to producing values that fall further from the mean. This is particularly useful for small samples.
  • The degrees of freedom (df) factor into its shape: fewer degrees of freedom result in thicker tails.
  • As sample size increases, the t-distribution approaches a normal distribution.
Some properties of the t-distribution: as you gather more data or your sample size increases, the distribution becomes more similar to the normal distribution, reducing the effect of its initial variability.
point estimate
In hypothesis testing, point estimates allow us to make an informed guess about population parameters based on sample statistics. Simply put, a point estimate is a single value estimate of a parameter and is often found using the sample means or proportions.
  • In the given exercise, the point estimate of the difference in milk consumption is calculated by subtracting the national mean from the sample mean. This calculation gives us a point estimate of 2.5 gallons.
  • This value indicates how much the sample data (Webster City's milk consumption) differs from the national average.
The closer a point estimate is to the actual parameter value, the more accurate our hypothesis test will be, which emphasizes the need for a reliable and representative sample.
This single number helps provide a clear, concise summary of the data that can be further explored with statistical testing.
critical value
The critical value in hypothesis testing helps determine whether to reject the null hypothesis. It is a threshold that our test statistic must exceed in order to indicate statistical significance.
  • In the context of the given problem, the critical value is determined using the t-distribution table considering the chosen significance level, most commonly denoted by \(\alpha\). For our problem, \(\alpha = 0.05\).
  • Using this level and the degrees of freedom (in our case, \(df = 15\) as the sample size is 16), the critical value is found to be approximately 1.753.
  • If the calculated t-statistic ( ≈ 2.083) exceeds this critical value, we have sufficient evidence to reject the null hypothesis.
Knowing the critical value before conducting tests gives us a clear benchmark to interpret whether the findings are statistically meaningful or could simply be due to random chance.
significance level
The significance level, often denoted by \(\alpha\), is a measure of the standard you set for evidence when deciding whether to reject a null hypothesis. Think of it as your threshold for how willing you are to risk making an error.

A significance level of \(\alpha = 0.05\) implies a 5% risk of concluding that a difference exists when there actually is none, called a Type I error. While choosing a significance level:
  • Lower values like \(\alpha = 0.01\) are used in more critical studies to minimize errors further.
  • Higher values like \(\alpha = 0.10\) can be used for preliminary analyses.
In this exercise, a 0.05 significance level was chosen, which means that if the probability of observing the test statistic under the null hypothesis is less than 5%, we reject \( H_0 \).
This provides a balance between being too cautious and making too many false discoveries.

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Most popular questions from this chapter

Nielsen reported that young men in the United States watch 56.2 minutes of prime-time TV daily (The Wall Street Journal Europe, November 18,2003 ). A researcher believes that young men in Germany spend more time watching prime- time TV. A sample of German young men will be selected by the researcher and the time they spend watching TV in one day will be recorded. The sample results will be used to test the following null and alternative hypotheses. \\[ \begin{array}{l} H_{0}: \mu \leq 56.2 \\ H_{\mathrm{a}}: \mu>56.2 \end{array} \\] a. What is the Type I error in this situation? What are the consequences of making this error? b. What is the Type II error in this situation? What are the consequences of making this error?

In 2001 , the U.S. Department of Labor reported the average hourly earnings for U.S. production workers to be \(\$ 14.32\) per hour (The World Almanac, 2003 ). A sample of 75 production workers during 2003 showed a sample mean of \(\$ 14.68\) per hour. Assuming the population standard deviation \(\sigma=\$ 1.45,\) can we conclude that an increase occurred in the mean hourly earnings since \(2001 ?\) Use \(\alpha=.05\).

The National Center for Health Statistics released a report that stated \(70 \%\) of adults do not exercise regularly (Associated Press, April 7, 2002). A researcher decided to conduct a study to see whether the claim made by the National Center for Health Statistics differed on a state-by-state basis. a. State the null and alternative hypotheses assuming the intent of the researcher is to identify states that differ from the \(70 \%\) reported by the National Center for Health Statistics. b. \(\quad\) At \(\alpha=.05,\) what is the research conclusion for the following states: Wisconsin: 252 of 350 adults did not exercise regularly California: 189 of 300 adults did not exercise regularly

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \geq 20 \\ H_{\mathrm{a}}: \mu<20 \end{array} \\] A sample of 50 provided a sample mean of \(19.4 .\) The population standard deviation is \(2 .\) a. Compute the value of the test statistic. b. What is the \(p\) -value? c. Using \(\alpha=.05,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?

Raftelis Financial Consulting reported that the mean quarterly water bill in the United States is \(\$ 47.50(U . S . \text { News \& World Report, August } 12,2002\) ). Some water systems are operated by public utilities, whereas other water systems are operated by private companies. An economist pointed out that privatization does not equal competition and that monopoly powers provided to public utilities are now being transferred to private companies. The concern is that consumers end up paying higher-than-average rates for water provided by private companies. The water system for Atlanta, Georgia, is provided by a private company. A sample of 64 Atlanta consumers showed a mean quarterly water bill of \(\$ 51\) with a sample standard deviation of \(\$ 12 .\) At \(\alpha=.05,\) does the Atlanta sample support the conclusion that above- average rates exist for this private water system? What is your conclusion?

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