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According to the most recent Bureau of Labor Statistics survey on time use in the United States, the average U.S. man spends \(67.20\) minutes per day eating and drinking. Suppose that a survey of 43 Norwegian men resulted in an average of \(81.10\) minutes per day eating and drinking [Note: This value is consistent with the data in a report by the Organization for Economic Cooperation and Development (Source: http://economix.blogs.nytimes.com/2009/05/05/obesity-and- the-fastness-of-food/)]. Assume that the population standard deviation for all Norwegian men is \(18.30\) minutes. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the average daily time spent eating and drinking by all Norwegian men is higher than \(67.20\) minutes. What is your conclusion at \(\alpha=.05\) ? b. Test the hypothesis of part a using the critical-value approach. Use \(\alpha=.01\).

Short Answer

Expert verified
For part a, reject or fail to reject the null hypothesis based on the comparison between the p-value and 0.05. For part b, reject or fail to reject the null hypothesis based on the comparison between the test statistic and the critical value for \(\alpha = 0.01\).

Step by step solution

01

State the Hypotheses

First, let's state the null hypothesis \(H_0\) and alternative hypothesis \(H_a\). \n\(H_0: \mu = 67.20\) (The mean (average) time Norwegian men spend eating and drinking per day is equal to the U.S average) \n\(H_a: \mu > 67.20\) (The mean time Norwegian men spend on eating and drinking per day is greater than the U.S. average)
02

Calculate Test Statistic

Now, calculate the z-score (test statistic) using the formula: \n\(z = \frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\) \nwhere \(\bar{X}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. \nSubstitute the given values into the formula: \n\(z = \frac{81.10 - 67.20}{\frac{18.30}{\sqrt{43}}}\)
03

Find P-value (part a)

Using a standard normal distribution table or a calculator, find the p-value associated with the calculated z-score. Note that this is an upper-tailed test, so the p-value is the area to the right of the calculated z-score.
04

Compare P-value with Significance Level (part a)

We now compare the p-value with the significance level \(\alpha = 0.05\). If the p-value is less than \(\alpha\), we reject the null hypothesis. If not, we fail to reject the null hypothesis.
05

Find Critical Value (part b)

For part b, we use the critical-value approach. For an alpha level \(\alpha = 0.01\), we find the critical value corresponding to this one-tailed test from the standard normal distribution table.
06

Compare Test Statistic with Critical Value (part b)

The sample data are in the rejection region if the test statistic is greater than the critical value. If so, reject the null hypothesis. If not, fail to reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often denoted as \(H_0\), is a statement assumed to be true unless evidence suggests otherwise. In hypothesis testing, it acts as the foundation of our investigation. Essentially, the null hypothesis is used to verify whether there is "no effect" or no difference in the context of our study.
In the given example, the null hypothesis \(H_0\) states that the average time Norwegian men spend eating and drinking is equal to the U.S. average of 67.20 minutes per day. This presumption implies any observed difference between the two might be due to random chance and not a significant deviation.
When conducting hypothesis testing, the null hypothesis is kept under scrutiny, and the hypothesis test will determine whether it can be rejected based on the data collected.
Alternative Hypothesis
The alternative hypothesis, denoted as \(H_a\), proposes what we suspect or hope to find true. It serves as a counter to the null hypothesis and represents a new perspective or idea about the data. In testing, validating the alternative hypothesis can lead to new insights or conclusions.
For our scenario, the alternative hypothesis \(H_a\) suggests that the average time Norwegian men spend eating and drinking is greater than 67.20 minutes per day. This hypothesis indicates that the Norwegian men, on average, spend more time eating and drinking compared to their U.S. counterparts.
  • This hypothesis is particularly useful when we want to see if there's an increase or a specific directional effect.
  • In tests like these, we aim to gather enough evidence to support \(H_a\) by rejecting the null hypothesis.
Understanding the alternative hypothesis helps frame the test as we focus on finding evidence to support this claim.
P-Value
The p-value is a crucial component in hypothesis testing, providing a way to measure the evidence against the null hypothesis. It indicates the probability of observing a test statistic as extreme as, or more extreme than, the one obtained from the sample data, assuming that the null hypothesis is true.
In the context of our exercise, once the z-score is calculated, the p-value is derived by finding the area to the right of this z-score on the standard normal distribution. Because we are conducting an upper-tailed test, we focus on this right-tail probability.
  • If the p-value is less than the chosen significance level \(\alpha\), often 0.05 or 0.01, it suggests that the null hypothesis is unlikely, hence we reject it.
  • The smaller the p-value, the stronger the evidence against the null hypothesis, suggesting the alternative hypothesis might be true.
The p-value helps determine whether the observed data is consistent with the null hypothesis or if there is sufficient evidence to support the alternative theory.
Critical-Value Approach
The critical-value approach is another technique to conduct hypothesis testing and involves comparing the test statistic against a threshold value, known as the critical value. This approach is grounded on the decision rule principle, where rejection or acceptance of the null is contingent upon whether the test statistic falls within a specific region of the distribution known as the rejection region.
For the exercise, given a significance level \(\alpha = 0.01\), we find the critical value corresponding to this one-tailed test from the standard normal distribution table. The threshold is determined by the point beyond which only 1% of the data would occur if the null hypothesis is true.
  • If the calculated test statistic exceeds the critical value, it implies the test statistic falls in the rejection region, leading us to reject the null hypothesis.
  • Conversely, if the test statistic does not exceed the critical value, we fail to reject the null hypothesis.
Utilizing the critical-value approach provides a clear-cut decision-making process by defining a fixed point in the distribution to judge the statistical significance of the results.

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Most popular questions from this chapter

A soft-drink manufacturer claims that its 12 -ounce cans do not contain, on average, more than 30 calories. A random sample of 64 cans of this soft drink, which were checked for calories, contained a mean of 32 calories with a standard deviation of 3 calories. Does the sample information support the alternative hypothesis that the manufacturer's claim is false? Use a significance level of \(5 \%\). Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value and \(\alpha=.05\) ?

The past records of a supermarket show that its customers spend an average of \(\$ 95\) per visit at this store. Recently the management of the store initiated a promotional campaign according to which each customer receives points based on the total money spent at the store, and these points can be used to buy products at the store. The management expects that as a result of this campaign, the customers should be encouraged to spend more money at the store. To check whether this is true, the manager of the store took a sample of 14 customers who visited the store. The following data give the money (in dollars) spent by these customers at this supermarket during their visits. \(\begin{array}{rrrrrrr}109.15 & 136.01 & 107.02 & 116.15 & 101.53 & 109.29 & 110.79 \\ 94.83 & 100.91 & 97.94 & 104.30 & 83.54 & 67.59 & 120.44\end{array}\) Assume that the money spent by all customers at this supermarket has a normal distribution. Using the \(5 \%\) significance level, can you conclude that the mean amount of money spent by all customers at this supermarket after the campaign was started is more than \(\$ 95 ?\) (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections \(3.1 .1\) and \(3.2 .2\) of Chapter \(3 .\) Then make the test of hypothesis about \(\mu .\) )

Consider the following null and alternative hypotheses: $$ H_{0}: p=.44 \text { versus } H_{1}: p<.44 $$ A random sample of 450 observations taken from this population produced a sample proportion of \(.39 .\) a. If this test is made at the \(2 \%\) significance level, would you reject the null hypothesis? Use the critical-value approach. b. What is the probability of making a Type I error in part a? c. Calculate the \(p\) -value for the test. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.01\) ? What if \(\alpha=.025\) ?

Consider the null hypothesis \(H_{0}: p=.65\). Suppose a random sample of 1000 observations is taken to perform this test about the population proportion. Using \(\alpha=.05\), show the rejection and nonrejection regions and find the critical value(s) of \(z\) for a a. left-tailed test b. two-tailed test c. right-tailed test

The Environmental Protection Agency (EPA) recommends that the sodium content in public water supplies should be no more than \(20 \mathrm{mg}\) per liter (http://www.disabledworld.com/artman/publish/ sodiumwatersupply.shtml). Forty samples were taken from a large reservoir, and the amount of sodium in each sample was measured. The sample average was \(23.5 \mathrm{mg}\) per liter. Assume that the population standard deviation is \(5.6 \mathrm{mg}\) per liter. The EPA is interested in knowing whether the average sodium content for the entire reservoir exceeds the recommended level. If so, the communities served by the reservoir will have to be made aware of the violation. a. Find the \(p\) -value for the test of hypothesis. Based on this \(p\) -value, would the communities need to be informed of an excessive average sodium level if the maximum probability of a Type I error is to be \(.05 ?\) What if the maximum probability of a Type I error is to be \(.01 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.05 .\) Would the communities need to be notified? What if \(\alpha=.01 ?\) What if \(\alpha\) is zero?

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