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For Exercises I through 25, perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Student Expenditures The average expenditure per student (based on average daily attendance) for a certain school year was \(\$ 10,337 with a population standard deviation of \)\$ 1560 . A survey for the next school year of 150 randomly selected students resulted in a sample mean of $\$ 10,798. Do these results indicate that the average expenditure has changed? Choose your own level of significance.

Short Answer

Expert verified
Expenditure per student has significantly changed, as shown by rejecting the null hypothesis.

Step by step solution

01

State the Hypotheses and Identify the Claim

We need to determine whether there has been a change in the average expenditure per student. Thus, we begin by setting up the null and alternative hypotheses. The null hypothesis \( H_0 \) is that the population mean \( \mu \) has not changed, i.e., \( H_0: \mu = 10,337 \). The alternative hypothesis \( H_a \) is that the population mean has changed, i.e., \( H_a: \mu eq 10,337 \). The claim is the alternative hypothesis that suggests the mean has changed.
02

Choose Level of Significance and Find Critical Value(s)

It is up to us to choose a level of significance; commonly, \( \alpha = 0.05 \) is used. Since this is a two-tailed test, we split \( \alpha \) into two, giving us \( \alpha/2 = 0.025 \). We use the standard normal distribution (z-distribution) because the population standard deviation is known. The critical values for a two-tailed test with \( \alpha = 0.05 \) are approximately \( z = \pm 1.96 \). This means if the test statistic falls outside this range, we will reject the null hypothesis.
03

Compute the Test Value

We calculate the test statistic using the formula: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] where \( \bar{x} = 10798 \) is the sample mean, \( \mu = 10337 \) is the population mean, \( \sigma = 1560 \) is the population standard deviation, and \( n = 150 \) is the sample size. Substituting these values results in: \[ z = \frac{10798 - 10337}{\frac{1560}{\sqrt{150}}} \approx 3.79 \]
04

Make the Decision

Since the calculated test value \( z \approx 3.79 \) falls outside the range of \( z = \pm 1.96 \), we reject the null hypothesis \( H_0 \). This indicates that there is significant evidence to suggest the mean expenditure per student has changed.
05

Summarize the Results

Based on the sample data and our calculations, we reject the null hypothesis and accept that there is a statistically significant change in the average expenditure per student. Therefore, the expenditure has indeed changed according to our hypothesis test with a significance level of 0.05.

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Key Concepts

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

Critical Values
Critical values are crucial components of hypothesis testing as they define the threshold at which you either accept or reject the null hypothesis. In hypothesis testing, once you have determined your desired level of significance, you use it to find the critical values. These values separate the distribution into regions where the null hypothesis would either be retained or rejected. When dealing with a two-tailed test, which is our scenario, you split your level of significance, usually denoted as \( \alpha \), into two equal parts, allocating half to each tail of the distribution.

Commonly, for a significance level of \( \alpha = 0.05 \), the critical values from the standard normal distribution (or z-distribution) are approximately \( z = \pm 1.96 \). These values are the cut-off points at which, if your test statistic falls into the extreme tail regions (beyond these cut-off points), you reject the null hypothesis.

Effectively, critical values create a boundary in the distribution. Any test statistic that lies beyond these boundaries offers enough evidence to discard the null hypothesis, suggesting that there's an effect or difference worth noting. Understanding how critical values work is essential for making informed decisions in statistical testing.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It essentially measures the number of standard errors that the sample mean is away from the population mean under the null hypothesis. When comparing the test statistic to the critical values, you make decisions about your hypotheses.

For the given problem about student expenditures, the test statistic is calculated using the z-score formula:
\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]
where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.

In this exercise, the calculation resulted in a test statistic of approximately 3.79. This means that the observed sample mean is 3.79 standard errors away from the claimed population mean, which indicates a significant deviation if we consider the standard normal distribution.

The test statistic is pivotal in hypothesis testing as it is the number used to determine whether the sample data provides enough evidence against the null hypothesis. Essentially, it connects the sample data to decision-making within the framework of hypothesis testing.
Null Hypothesis
The null hypothesis, often represented as \( H_0 \), is a fundamental aspect of hypothesis testing. It is a statement that suggests there is no effect or no difference, and it serves as the default or original assumption that you aim to test. The significance of \( H_0 \) is that it establishes a baseline for statistical testing.

In the context of the exercise, the null hypothesis posits that the population mean, \( \mu \), has not changed and is equal to 10,337. This forms the premise upon which you will conduct statistical evaluation. The null hypothesis functions as the statement to be tested against the evidence provided by the sample data.

A vital part of hypothesis testing involves either rejecting or failing to reject the null hypothesis based on the test statistic and critical values. If the test statistic falls within the critical region, as determined by the critical values, you reject the null hypothesis. Otherwise, there isn't sufficient evidence to discard it, and you "fail to reject" it. Importantly, failing to reject the null hypothesis doesn't prove it true, it simply indicates that there wasn't enough evidence against it.
Two-Tailed Test
A two-tailed test is one of the approaches used in hypothesis testing to determine if there is a statistically significant difference in either direction from the hypothesized parameter. This type of test is applicable when you are interested in identifying departures in either direction (either higher or lower than the population mean or specified value).

In the school expenditure exercise, the hypothesis tests whether there has been a change, either an increase or decrease, in the mean expenditure. This means deviations in both directions are of interest, and thus a two-tailed test is appropriate.

Utilizing a two-tailed test involves splitting the significance level \( \alpha \) equally across both tails of the distribution. For instance, with a typical \( \alpha = 0.05 \), each tail would have 0.025. Critical values thus play an essential role in determining the rejection regions in both tails.

A conclusion is reached based on whether the test statistic falls within either tail; if it does, then it stands as evidence against the null hypothesis, suggesting that the deviation from the hypothesized value is sufficiently significant.

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Most popular questions from this chapter

In hypothesis testing, why can't the hypothesis be proved true?

For Exercises 5 through \(20,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Runaways A corrections officer read that \(58 \%\) of runaways are female. He believes that the percentage is higher than \(58 .\) He selected a random sample of 90 runaways and found that 63 were female. At \(\alpha=0.05,\) can you conclude that his belief is correct?

Prison Time According to a public service website, $$69.4 \%$$ of white collar criminals get prison time. A randomly selected sample of 165 white collar criminals revealed that 120 were serving or had served prison time. Using $$\alpha=0.05,$$ test the conjecture that the proportion of white collar criminals serving prison time differs from $$69.4 \%$$ in two different ways.

For Exercises 5 through \(20,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Stocks and Mutual Fund Ownership It has been found that \(50.3 \%\) of U.S. households own stocks and mutual funds. A random sample of 300 heads of households indicated that 171 owned some type of stock. At what level of significance would you conclude that this was a significant difference?

For Exercises 7 through \(23,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. Cell Phone Bills The average monthly cell phone bill was reported to be $\$ 50.07 by the U.S. Wireless Industry. Random sampling of a large cell phone company found the following monthly cell phone charges (in dollars): $$\begin{array}{llll}{55.83} & {49.88} & {62.98} & {70.42} \\ {60.47} & {52.45} & {49.20} & {50.02} \\ {58.60} & {51.29}\end{array}$$ At the 0.05 level of significance, can it be concluded that the average phone bill has increased?

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