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Chapter 16: Problem 563

A sample of size 49 yielded the values \(\bar{x}=87.3\) and \(\mathrm{s}^{2}=162\). Test the hypothesis that \(\mu=95\) versus the alternative that it is less. Let \(\alpha=.01\).

Short Answer

Expert verified
We conducted a one-tailed t-test with a significance level of \(\alpha = 0.01\) to test the hypothesis that the population mean \(\mu = 95\) versus the alternative that it is less. The test statistic was approximately \(t \approx -2.96\), and the critical value was \(t_{0.01, 48} \approx -2.41\). Since the test statistic is less than the critical value, we reject the null hypothesis and have enough evidence to support the alternative hypothesis that the population mean is less than 95.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis (\(H_0\)) is that the population mean \(\mu\) is equal to 95, and the alternative hypothesis (\(H_a\)) is that the population mean is less than 95: \[H_0: \mu = 95\] \[H_a: \mu < 95\]
02

Calculate the test statistic

We are given the sample mean \(\bar{x} = 87.3\), sample variance \(s^2 = 162\), and sample size \(n = 49\). The test statistic for a one-sample t-test is calculated as: \[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\] Plugging in the given values: \[t = \frac{87.3 - 95}{\sqrt{162}/\sqrt{49}}\] \[t = \frac{-7.7}{\sqrt{162}/7}\] Calculate the test statistic: \[t \approx -2.96\]
03

Find the critical value

We are given a significance level of \(\alpha = 0.01\). Since this is a one-tailed test, we will find the critical value corresponding to this significance level with \(n-1\) degrees of freedom. Using a t-distribution table or calculator, we find that the critical value is approximately: \[t_{0.01, 48} \approx -2.41\]
04

Compare the test statistic and critical value

Now we will compare the test statistic with the critical value: \[-2.96 < -2.41\] Since the test statistic is less than the critical value, we reject the null hypothesis.
05

Conclusion

Based on the results of the one-tailed t-test with a significance level of \(\alpha = 0.01\), we reject the null hypothesis that the population mean \(\mu = 95\). Instead, we have enough evidence to support the alternative hypothesis that the population mean is less than 95.

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Key Concepts

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

Understanding the One-Sample t-Test
The one-sample t-test is a statistical method used to determine if the mean of a single sample is significantly different from a known or hypothesized population mean. It's particularly useful when the population standard deviation is unknown and the sample size is relatively small. In the exercise provided, we're testing whether our sample mean of 87.3 is significantly less than the population mean of 95.

Here's how it works:
  • Formulate Hypotheses: Begin by stating the null hypothesis (\(H_0: \mu = 95\)) and the alternative hypothesis (\(H_a: \mu < 95\)). The null hypothesis assumes no effect or difference, and the alternative hypothesis suggests a deviation.

  • Calculate the Test Statistic: The test statistic is calculated as \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\). It tells us how many standard deviations our sample mean is from the hypothesized population mean.

This test helps us make decisions about our data compared to a known value.
Critical Value and Its Role
In hypothesis testing, the critical value serves as a cutoff point that helps determine whether to reject the null hypothesis. This value comes from a statistical distribution (like the t-distribution in our example) and represents the threshold beyond which a test statistic is considered too extreme.

For the given exercise:
  • Find the Critical Value: We use a t-distribution table or calculator, considering the significance level (\(\alpha = 0.01\)) and degrees of freedom (\(n-1 = 48\)).

  • Interpreting the Critical Value: Here, the critical value is \(t_{0.01, 48} \approx -2.41\). This value signifies the point beyond which any test statistic would lead to rejecting the null hypothesis.

Comparing the test statistic and critical value helps us make informed decisions in hypothesis testing.
The Significance Level Explained
When conducting hypothesis tests, the significance level (denoted by \(\alpha\)) plays a crucial role. It represents the probability of rejecting the null hypothesis when it is actually true, often setting the threshold for making statistical decisions.

In this example, the significance level is set to \(0.01\) (or 1%), indicating that we are willing to accept a 1% risk of incorrectly rejecting the null hypothesis.

Here's why it matters:
  • Defines Confidence: A lower \(\alpha\) value, like 0.01, means we require stronger evidence to reject the null hypothesis compared to a higher \(\alpha\), such as 0.05.

  • Guides Decision Making: It helps in interpreting results. If our calculated test statistic falls into the "rejection region," determined by \(\alpha\), we reject the null hypothesis.

Overall, the significance level helps quantify and manage the risk of drawing incorrect conclusions from statistical tests.

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Most popular questions from this chapter

In a recent science fiction novel, aliens invented a death ray that killed 3 Earthman. The lethal doses were 10,11 , and 12 units of the ray. The mean lethal dose for Moonmen is \(13.30\) units. For this sample of 3 Earthmen \(\underline{\mathrm{X}}=11\) and \(\mathrm{S}=1.0\) and a test of significance of the difference between the mean for Earthmen and Moonmen yields \(\mathrm{t}=3.98\), not significant at the \(\alpha=.05\) level of significance. Then 6 more Earthmen were killed with lethal doses of \(9,10,11,12,13\) and \(14 .\) For this sample, \(\underline{\mathrm{X}}=11.5\), and \(\mathrm{S}=1.87\), and a test of significance of the difference between the mean for Earthmen and Moonmen yields \(\mathrm{t}=2.36\), also not significant at \(\alpha=.05\). What can be said if we combine the results of these two samples?

It is believed that no more than \(40 \%\) of the students in a certain college wear glasses. Of 64 students surveyed, 40 or \(62.5 \%\) are found to wear glasses. Does this contradict the belief that no more than \(40 \%\) of the students wear glasses? Use a \(5 \%\) level of sionificance.

Suppose we have a binomial distribution for which \(\mathrm{H}_{0}\) is that the probability of success on a single trial, \(\mathrm{p}=1 / 2\). Suppose also that \(\mathrm{H}_{1}\) is \(\mathrm{p}=2 / 3\). Show how the power of the normal approximation to the binomial test increases as n increases by finding the critical value, \(K\), and the type II error, \(\beta\), for each of the following values of n: \(36,64,100,144\), and 196 . For which of these values of \(\mathrm{N}\) does \(\beta\) first fall as low as \(.5 ?\) Use \(\alpha=.01\).

A man has just purchased a trick die which was advertised as not yielding the proper proportion of sixes. He wonders whether the advertising was correct, and tests the advertising claim by rolling the die 100 times. The 100 rolls yielded ten sixes. Should he conclude that the advertising was legitimate?

The Selective Service director of a certain state suspects that the proportion of men from urban areas who are physically unfit for military service is more than 5 percentage points greater than the proportion of physically unfit men from rural areas. He decides to treat the men called for a physical examination from urban areas during the next month as a random sample from a binomial population, and those from the rural areas as a random sample from a second binomial population. During the next month 3214 men were called from urban areas and 2011 from rural areas. There were 1078 physical rejects from the urban areas and 543 from rural areas. Formulate the appropriate null and alternative hypotheses, and test the null hypothesis at the \(\alpha=.05\) level.
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