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According to the Center on Budget and Policy Priorities' article "Curbing Flexible Spending Accounts Could Help Pay for Health Care Reform" (revised June \(10,2009),\) flexible-spending accounts encourage the over consumption of health care. People buy things they do not need; otherwise they lose the money. In 2007 , for those who did not use all of their account (about one out of every seven), the average amount lost was \(\$ 723\) Suppose a random sample of 150 employees who did not use all of their funds in 2009 is taken and an average amount of \(\$ 683\) was lost. Test the hypothesis that there is no significant difference in the average amount forfeited. Assume that \(\sigma=\$ 307\) per year. Use \(\alpha=0.05\) a. Define the parameter. b. State the null and alternative hypotheses. c. Specify the hypothesis test criteria. d. Present the sample evidence. e. Find the probability distribution information. f. Determine the results.

Short Answer

Expert verified
The test follows the straightforward steps of hypothesis testing: defining the parameter, stating the hypotheses, testing criteria, presenting evidence and results. The final step depends on the Z score calculated.

Step by step solution

01

Define the parameter

The parameter we are trying to estimate is the population mean (\(饾渿\)) of money forfeited from flexible spending accounts.
02

State the null and alternative hypotheses

The null hypothesis (\(H_0\)) is that the population mean is the same in 2009 as it was in 2007 (\(饾渿 = \$723\)). The alternative hypothesis (\(H_1\)) is that the population mean is not the same in 2009 as it was in 2007 (\(饾渿 鈮 \$723\)).
03

Specify the hypothesis test criteria

We will use a z-test for this hypothesis testing since we have the standard deviation of the population. The test statistic is given by \(Z = \frac{(X - 饾渿_0)}{(饾湈/鈭歯)}\) where X is the sample mean, 饾渿_0 is the population mean under the null hypothesis, 饾湈 is the standard deviation of the population and n is the sample size. The critical value for a two-tailed test at \(\alpha = 0.05\) is 卤1.96.
04

Present the sample evidence

A random sample of 150 employees was taken where the sample mean (X) is \$683. The sample size (n) is 150.
05

Find the probability distribution information

The standard deviation of the population (\(饾湈\)) is \$307. Substituting these values into the formula for the test statistic, we get \(Z = \frac{(\$683 - \$723)}{(\$307/鈭150)} \). Compute this to get the Z score.
06

Determine the results

Compare the calculated Z score with the critical value. If the Z score is greater than 1.96 or less than -1.96, we reject the null hypothesis. Otherwise, we do not reject the null hypothesis.

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Key Concepts

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

Z-test
The Z-test is a type of hypothesis test used when the population standard deviation is known and the sample size is large. It is commonly used to determine if there is a significant difference between the population mean and a sample mean. The Z-test helps to assess whether any observed differences are due to random chance or if they are statistically significant. In the context of this exercise, a Z-test enables us to compare the average amount of money lost in flexible spending accounts in 2009 to that in 2007.
  • Standard deviation is known
  • Sample size is large (usually n > 30)
  • Assumes normal distribution
The formula for the Z-test is: \[ Z = \frac{(X - \mu_0)}{(\sigma/\sqrt{n})} \]where \(X\) is the sample mean, \(\mu_0\) is the population mean under the null hypothesis, \(\sigma\) is the standard deviation of the population, and \(n\) is the sample size. This formula shows how many standard deviations the sample mean is from the population mean. A higher Z-score indicates a higher deviation from what is expected, leading to a potential rejection of the null hypothesis.
Null Hypothesis
The null hypothesis, often represented as \(H_0\), is a statement that there is no effect or no difference. It serves as the default or starting assumption in hypothesis testing. In the exercise, the null hypothesis is that there is no difference in the average amount forfeited from flexible spending accounts between 2007 and 2009. Mathematically, this is expressed as: \[ H_0: \mu = \$723 \]where \(\mu\) represents the population mean in 2009. By testing the null hypothesis, researchers can determine if any observed differences in the data are statistically significant or if they might have occurred by random chance. Rejecting the null hypothesis suggests that there is evidence against it, supporting the idea that the population mean might be different from 723.
Alternative Hypothesis
The alternative hypothesis, denoted as \(H_1\) or \(H_a\), is the statement that opposes the null hypothesis. It represents what the researcher aims to demonstrate. In this exercise, the alternative hypothesis suggests that the average amount forfeited in 2009 is different from the amount in 2007. Therefore, it is expressed as:\[ H_1: \mu eq \$723 \]This two-tailed alternative hypothesis is used when we are interested in detecting any difference from the hypothesized mean, whether it is higher or lower. If the hypothesis test leads to the rejection of the null hypothesis, it provides support for the alternative hypothesis, indicating a significant change in the average amounts forfeited from the flexible spending accounts between the two years.
Population Mean
The population mean, denoted as \(\mu\), is the average of a set of values for a whole population. In many real-world scenarios, including this exercise, the population mean can be a theoretical value that we can only estimate through sampling due to the large size or inaccessibility of the population. Here, the parameter of interest is the population mean of money forfeited from flexible spending accounts.
  • A theoretical average of the entire data set
  • Often estimated through sample data
  • Different from the sample mean, which is specific to one sample
The task within this hypothesis test is to determine if \(\mu\) in 2009 could be the same as \$723, the population mean in 2007, or if there has been a significant change. This value forms the basis of the null hypothesis, serving as a point of comparison for the sample mean obtained from the 2009 data.
Standard Deviation
Standard deviation, represented by the symbol \(\sigma\), is a measure of the dispersion or spread of a set of values around the mean. A smaller standard deviation indicates that the values tend to be close to the mean, while a larger standard deviation shows more variability. In hypothesis testing, knowing the standard deviation of the population is critical for calculating the Z-test statistic.
  • Measure of spread in data
  • Population standard deviation is used in Z-test
In this particular exercise, the population standard deviation of the amount forfeited is \$307. This value helps determine how much variation there is in the money forfeited by employees. The standard deviation is used in the Z-test formula to assess how far the sample mean is from the hypothesized population mean, helping us decide whether to accept or reject the null hypothesis. Understanding this metric is crucial for interpreting the results of the hypothesis test accurately.

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Most popular questions from this chapter

Suppose you want to test the hypothesis that "the mean salt content of frozen 'lite' dinners is more than \(350 \mathrm{mg}\) per serving." An average of \(350 \mathrm{mg}\) is an acceptable amount of salt per serving; therefore, you use it as the standard. The null hypothesis is "The average content is not more than \(350 \mathrm{mg} "(\mu=350) .\) The alternative hypothesis is "The average content is more than \(350 \mathrm{mg} "\) \((\mu > 350)\) a. Describe the conditions that would exist if your decision results in a type I error. b. Describe the conditions that would exist if your decision results in a type II error.

From a population of unknown mean \(\mu\) and a standard deviation \(\sigma=5.0,\) a sample of \(n=100\) is selected and the sample mean 41.5 is found. Compare the concepts of estimation and hypothesis testing by completing the following: a. Determine the \(95 \%\) confidence interval for \(\mu\) b. Complete the hypothesis test involving \(H_{a}: \mu \neq 40\) using the \(p\) -value approach and \(\alpha=0.05\) c. Complete the hypothesis test involving \(H_{a}: \mu \neq 40\) using the classical approach and \(\alpha=0.05\) d. On one sketch of the standard normal curve, locate the interval representing the confidence interval from part a; the \(z \star, p\) -value, and \(\alpha\) from part \(b ;\) and the \(z \\#\) and critical regions from part c. Describe the relationship between these three separate procedures.

Suppose we want to test the hypothesis that the mean hourly charge for automobile repairs is at least \(\$ 60\) per hour at the repair shops in a nearby city. Explain the conditions that would exist if we make an error in decision by committing a type I error. What about a type II error?

a. If \(\alpha\) is assigned the value \(0.001,\) what are we saying about the type I error? b. If \(\alpha\) is assigned the value \(0.05,\) what are we saying about the type I error? c. If \(\alpha\) is assigned the value \(0.10,\) what are we saying about the type I error?

The calculated value of the test statistic is actually the number of standard errors that the sample mean differs from the hypothesized value of \(\mu\) in the null hypothesis. Suppose that the null hypothesis is \(H_{o}: \mu=4.5, \sigma\) is known to be \(1.0,\) and a sample of size 100 results in \(\bar{x}=4.8\) a. How many standard errors is \(\bar{x}\) above \(4.5 ?\) b. If the alternative hypothesis is \(H_{a}: \mu>4.5\) and \(\alpha=0.01,\) would you reject \(H_{o} ?\)

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