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91Ó°ÊÓ

A magazine publisher tells potential advertisers that the mean household income of its regular readership is $$\$ 61,500.$$ An advertising agency wishes to test this claim against the alternative that the mean is smaller. A sample of 40 randomly selected regular readers yields mean income $$\$ 59,800$$ with standard deviation $$\$ 5,850.$$ Perform the relevant test at the \(1 \%\) level of significance.

Short Answer

Expert verified
We do not reject the null hypothesis; the evidence is insufficient to say the mean income is less than $61,500.

Step by step solution

01

Define Hypotheses

Begin by defining the null and alternative hypotheses. The null hypothesis \((H_0)\) claims that the mean household income is \(\mu = 61,500\), while the alternative hypothesis \((H_a)\) claims that the mean income is smaller, \(\mu < 61,500\).
02

Determine Significance Level

The level of significance \(\alpha\) is given as \(0.01\) or \(1\%\). This is the probability of rejecting the null hypothesis when it is actually true.
03

Calculate Test Statistic

Use the formula for the test statistic in a t-test: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \] where \(\bar{x} = 59,800\), \(\mu = 61,500\), \(s = 5,850\), and \(n = 40\). Substituting these values gives: \[ t = \frac{59,800 - 61,500}{5,850/\sqrt{40}} \approx -1.84 \].
04

Find Critical Value

Look up the critical t-value for a one-tailed test at \(1\%\) significance level with \(n-1 = 39\) degrees of freedom using a t-distribution table, which is approximately \(-2.426\).
05

Make a Decision

Compare the calculated test statistic \(-1.84\) to the critical t-value \(-2.426\). Since \(-1.84\) is greater than \(-2.426\), we do not reject the null hypothesis.
06

Conclusion

Since we did not reject the null hypothesis, there is insufficient evidence to support the claim that the mean household income of the magazine's readership is less than \(\$61,500\) at the \(1\%\) significance level.

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Key Concepts

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

Null Hypothesis
In the world of hypothesis testing, the null hypothesis plays an integral role. It represents the default or initial assumption that there is no effect or no difference in the situation being analyzed. In our example, the null hypothesis, denoted as \(H_0\), assumes that the mean household income for the magazine's readership is indeed \(\mu = 61,500\). This assumption helps provide a baseline for statistical testing.

When conducting a hypothesis test, the goal is to gather enough evidence to determine whether this hypothesis can be rejected in favor of an alternative. If the data strongly contradicts the null hypothesis, then researchers may find justification for dismissing it. However, remember that failing to reject the null hypothesis doesn't prove it true; it simply indicates a lack of evidence against it.
t-test
A t-test is a statistical method used to compare a sample mean to a known value, or between two groups. In this exercise, we use a one-sample t-test to determine if the average income of the readership differs from the claimed value. This test is applicable when the sample size is relatively small and the population standard deviation is unknown, which is often the case in real-world scenarios.

To conduct the t-test, we calculate a test statistic using the formula:
  • \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\)
  • \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
This calculated \(t\) value tells us how far our sample mean is from the population mean in terms of standard errors. A large absolute value suggests a significant difference.
Alternative Hypothesis
The alternative hypothesis offers a contrasting view to the null hypothesis. In this exercise, we have an alternative hypothesis, \(H_a\), which posits that the mean household income is less than \(61,500\). This hypothesis reflects our actual research question or suspicion that challenges the status quo set by the null hypothesis.

When testing, researchers look for evidence to support the alternative hypothesis. It's essentially what you're "trying to prove" with your data. If the test statistic falls into a critical region or surpasses a threshold, then we gain support for the alternative over the null hypothesis.

Thus, in our example, strong evidence for \(H_a\) would suggest the magazine's readership might not be as affluent as initially claimed.
Significance Level
The significance level, denoted by \(\alpha\), helps in determining the threshold for rejecting the null hypothesis. It's the probability of making a Type I error – rejecting \(H_0\) when it is true.

Commonly expressed as a percentage, the lower the significance level, the stricter the criteria for rejection, reducing the chance of a false positive.

For this problem, a \(1\%\) significance level is chosen. Such a low \(\alpha\) implies a rigorous test where findings must be extremely conclusive before dismissing the null hypothesis. In practice, a particular \(\alpha\) level should align with the acceptable risk for the specific field of study, balancing the likelihood of errors with the importance of the decision.

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Most popular questions from this chapter

The recommended daily calorie intake for teenage girls is 2,200 calories/day. A nutritionist at a state university believes the average daily caloric intake of girls in that state to be lower. Test that hypothesis, at the \(5 \%\) level of significance, against the null hypothesis that the population average is 2,200 calories/day using the following sample data: \(n=36, x-2,150, s=203\).

Compute the value of the test statistic for the indicated test, based on the information given. a. Testing \(H_{0}: \mu=342\) vs. \(H a: \mu<342, \sigma=11.2, n=40, x-=339, s=10.3\) b. Testing \(H_{0}: \mu=105\) vs. Ha: \(\mu>105, \sigma=5.3, n=80, x-=107, s=5.1\) c. Testing \(H_{0}: \mu=-13.5\) vs. Ha: \(\mu \neq-13.5, \sigma\) unknown, \(n=32, x-=-13.8, s=1.5\) d. Testing \(H_{0}: \mu=28\) vs. \(H a: \mu \neq 28, \sigma\) unknown, \(n=68, x-27.8, s=1.3\)

The target temperature for a hot beverage the moment it is dispensed from a vending machine is \(170^{\circ} \mathrm{F}\). A sample of ten randomly selected servings from a new machine undergoing a pre- shipment inspection gave mean temperature \(173^{\circ} \mathrm{F}\) with sample standard deviation \(6.3^{\circ} \mathrm{F}\). a. Assuming that temperature is normally distributed, perform the test that the mean temperature of dispensed beverages is different from \(170^{\circ} \mathrm{F}\), at the \(10 \%\) level of significance. b. The sample mean is greater than 170 , suggesting that the actual population mean is greater than \(170^{\circ} \mathrm{F}\). Perform this test, also at the \(10 \%\) level of significance. (The computation of the test statistic done in part (a) still applies here.)

A dairy farm uses the somatic cell count (SCC) report on the milk it provides to a processor as one way to monitor the health of its herd. The mean SCC from five samples of raw milk was 250,000 cells per milliliter with standard deviation 37,500 cell/ml. Test whether these data provide sufficient evidence, at the \(10 \%\) level of significance, to conclude that the mean SCC of all milk produced at the dairy exceeds that in the previous report, 210,250 cell/ml. Assume a normal distribution of SCC.

In the past the average length of an outgoing telephone call from a business office has been 143 seconds. A manager wishes to check whether that average has decreased after the introduction of policy changes. A sample of 100 telephone calls produced a mean of 133 seconds, with a standard deviation of 35 seconds. Perform the relevant test at the \(1 \%\) level of significance.

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