/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 The following data are based on ... [FREE SOLUTION] | 91Ó°ÊÓ

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The following data are based on information taken from the book Navajo Architecture: Forms, History, Distributions, by S. C. Jett and V. E. Spencer (University of Arizona Press). A survey of houses and traditional hogans was made in a number of different regions of the modern Navajo Indian Reservation. The following table is the result of a random sample of eight regions on the Navajo Reservation. \(\begin{array}{lcc} \hline \begin{array}{l} \text { Area on } \\ \text { Navajo Reservation } \end{array} & \begin{array}{c} \text { Number of } \\ \text { Inhabited Houses } \end{array} & \begin{array}{c} \text { Number of } \\ \text { Inhabited Hogans } \end{array} \\ \hline \text { Bitter Springs } & 18 & 13 \\ \text { Rainbow Lodge } & 16 & 14 \\ \text { Kayenta } & 68 & 46 \\ \text { Red Mesa } & 9 & 32 \\ \text { Black Mesa } & 11 & 15 \\ \text { Canyon de Chelly } & 28 & 47 \\ \text { Cedar Point } & 50 & 17 \\ \text { Burnt Water } & 50 & 18 \\ \hline \end{array}\) Does this information indicate that the population mean number of inhabited houses is greater than that of hogans on the Navajo Reservation? Use a \(5 \%\) level of significance.

Short Answer

Expert verified
Yes, the data suggest that the mean number of inhabited houses is greater than the number of hogans.

Step by step solution

01

Formulating Hypotheses

We start by setting up our null and alternative hypotheses. The null hypothesis, \(H_0\), states that the mean number of inhabited houses, \(\mu_1\), is equal to the mean number of inhabited hogans, \(\mu_2\). The alternative hypothesis, \(H_a\), asserts that \(\mu_1\) is greater than \(\mu_2\). Mathematically, these are expressed as: \(H_0: \mu_1 = \mu_2\) and \(H_a: \mu_1 > \mu_2\).
02

Calculate Sample Means and Differences

For each region, calculate the difference between the number of inhabited houses and hogans. Then, find the sample mean of these differences: \(d_i = \text{houses} - \text{hogans}\). Calculate the mean of these differences \(\bar{d}\) across all eight regions.
03

Compute Standard Deviation of Differences

Calculate the standard deviation \(s_d\) of the differences \(d_i\) from Step 2. This measures the variability in the differences between houses and hogans across the regions.
04

Conduct t-test for Paired Sample

We use a paired t-test to determine if the mean difference \(\bar{d}\) is significantly greater than zero. This involves the formula: \( t = \frac{\bar{d}}{s_d/\sqrt{n}} \), where \(n\) is the number of regions (which is 8). Compute the t-statistic using the values from Steps 2 and 3.
05

Determine Critical Value and Compare

At a 5% level of significance and based on a one-tailed test with \(n-1=7\) degrees of freedom, find the critical value from the t-distribution table. Compare the computed t-statistic from Step 4 with this critical value.
06

Conclusion

If the computed t-statistic is greater than the critical value, reject the null hypothesis \(H_0\) and accept the alternative hypothesis \(H_a\). This would indicate that the mean number of inhabited houses is indeed greater than that of hogans on the Navajo Reservation.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a powerful statistical method used to make decisions or inferences about population parameters based on sample data. It begins with stating two hypotheses:
- The **null hypothesis** \(H_0\) is a statement of no effect or no difference. In our case, it claims that the average number of inhabited houses on the Navajo Reservation is the same as the average number of inhabited hogans.
- The **alternative hypothesis** \(H_a\), suggests the opposite of the null. Here, it asserts that the average number of houses is greater than that of hogans.

Once the hypotheses are defined, we collect sample data to test them. Statistical tests, like the paired t-test, help determine whether we have enough evidence to reject the null hypothesis in favor of the alternative. This decision is not made lightly, as it involves comparing calculated test statistics to critical values determined by the chosen level of significance. If the test statistic exceeds the critical value, it provides evidence to reject \(H_0\) and accept \(H_a\).
Sample Mean
The sample mean is a fundamental statistic that estimates the central value of a sample data set. It helps in understanding the average behavior of observations.
In our analysis, we calculated the sample mean of the differences between the number of inhabited houses and hogans.- First, find the difference between houses and hogans for each region \(d_i = ext{houses} - ext{hogans}\).- Then, compute the average of these differences to get the sample mean \(\overline{d}\).
This average difference helps us assess whether houses consistently outnumber hogans across the sampled regions. The sample mean is crucial in the paired t-test, representing the observed effect size we evaluate against the null hypothesis.
Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are around the mean. In our study, we compute the standard deviation of the differences between the number of inhabited houses and hogans. This gives us an idea of the variability or spread of these differences across different regions.
Steps include:
  • Subtract the sample mean difference from each individual difference.

  • Square these outcomes to eliminate negative values.

  • Compute the average of these squared differences.

  • Finally, take the square root of this average to obtain the standard deviation \(s_d\).

Understanding standard deviation is vital since it helps in calculating the test statistic for the paired t-test. A smaller standard deviation indicates that the differences are consistently near the average, providing stronger evidence against the null hypothesis if a significant effect is detected.
Level of Significance
The level of significance is a threshold set by the researcher, which determines how extreme the test statistic must be before we can reject the null hypothesis. It is denoted by \( \alpha \), and commonly set at 0.05, 0.01, or 0.10.

In our scenario, we used a 5% level of significance \(\alpha = 0.05\). This means if the probability of observing the test statistic is less than 5%, given the null hypothesis is true, we reject \(H_0\).
This chosen level is crucial because it balances the risk of two types of errors:
  • **Type I error**: Rejecting a true null hypothesis.
  • **Type II error**: Failing to reject a false null hypothesis.

By carefully selecting \(\alpha\), researchers can control and minimize the chances of making incorrect decisions based on sample data.

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Most popular questions from this chapter

A random sample of \(n_{1}=16\) communities in western Kansas gave the following information for people under 25 years of age. \(x_{1}:\) Rate of hay fever per 1000 population for people under 25 \(\begin{array}{rrrrrrrr}98 & 90 & 120 & 128 & 92 & 123 & 112 & 93 \\ 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88\end{array}\) A random sample of \(n_{2}=14\) regions in western Kansas gave the following information for people over 50 years old. \(x_{2}\) : Rate of hay fever per 1000 population for people over 50 \(\begin{array}{llllllr}95 & 110 & 101 & 97 & 112 & 88 & 110 \\ 79 & 115 & 100 & 89 & 114 & 85 & 96\end{array}\) (Reference: National Center for Health Statistics.) i. Use a calculator to verify that \(\bar{x}_{1} \approx 109.50, s_{1} \approx 15.41, \bar{x}_{2} \approx 99.36\), and \(s_{2} \approx 11.57\) ii. Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use \(\alpha=0.05\).

Alisha is conducting a paired differences test for a "before \((B\) score) and after \((A\) score \() "\) situation. She is interested in testing whether the average of the "before" scores is higher than that of the "after" scores. (a) To use a right-tailed test, how should Alisha construct the differences between the "before" and "after" scores? (b) To use a left-tailed test, how should she construct the differences between the "before" and "after" scores?

In the article cited in Problem 15, the results of the following experiment were reported. Form 2 of the Gates-MacGintie Reading Test was administered to both an experimental group and a control group after 6 weeks of instruction, during which the experimental group received peer tutoring and the control group did not. For the experimental group \(n_{1}=30\) children, the mean score on the vocabulary portion of the test was \(\bar{x}_{1}=368.4\), with sample standard deviation \(s_{1}=39.5\). The average score on the vocabulary portion of the test for the \(n_{2}=30\) subjects in the control group was \(\bar{x}_{2}=349.2\), with sample standard deviation \(s_{2}=56.6 .\) Use a \(1 \%\) level of significance to test the claim that the experimental group performed better than the control group.

In general, if sample data are such that the null hypothesis is rejected at the \(\alpha=1 \%\) level of significance based on a two-tailed test, is \(H_{0}\) also rejected at the \(\alpha=1 \%\) level of significance for a corresponding one-tailed test? Explain.

Consider a hypothesis test of difference of proportions for two independent populations. Suppose random samples produce \(r_{1}\) successes out of \(n_{1}\) trials for the first population and \(r_{2}\) successes out of \(n_{2}\) trials for the second population. What is the best pooled estimate \(\bar{p}\) for the population probability of success using \(H_{0}: p_{1}=p_{2} ?\)

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