91Ó°ÊÓ

Individuals filing federal income tax returns prior to March 31 received an average refund of \(\$ 1056 .\) Consider the population of "last-minute" filers who mail their tax return during the last five days of the income tax period (typically April 10 to April 15 ). a. A researcher suggests that a reason individuals wait until the last five days is that on average these individuals receive lower refunds than do early filers. Develop appropriate hypotheses such that rejection of \(H_{0}\) will support the researcher's contention. b. For a sample of 400 individuals who filed a tax return between April 10 and \(15,\) the sample mean refund was \(\$ 910 .\) Based on prior experience a population standard deviation of \(\sigma=\$ 1600\) may be assumed. What is the \(p\) -value? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.

Step by step solution

01

Define Hypotheses

To test whether last-minute filers receive lower refunds compared to early filers, we set up the hypotheses as follows: \( H_0: \mu = 1056 \) (Last-minute filers have the same average refund as early filers) and \( H_1: \mu < 1056 \) (Last-minute filers have a lower average refund than early filers).

02

Calculate the Test Statistic

Use the test statistic formula for the mean: \( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \). Here, \( \bar{x} = 910 \), \( \mu = 1056 \), \( \sigma = 1600 \), and \( n = 400 \). Substituting in these values, \( z = \frac{910 - 1056}{\frac{1600}{\sqrt{400}}} = \frac{-146}{80} = -1.825 \).

03

Find the p-value

The p-value is determined from the standard normal distribution table for the calculated \( z \)-value of \(-1.825\). The p-value corresponding to \(-1.825\) is approximately 0.0340.

04

Decision Using p-value Approach

At \( \alpha = 0.05 \), compare the p-value (0.0340) to \( \alpha \). Since the p-value is less than \( \alpha \), we reject \( H_0 \). This supports the researcher's contention that last-minute filers receive lower refunds.

05

Critical Value Approach

Find the critical z-value for \( \alpha = 0.05 \) in a one-tailed test. From the z-table, the critical value is \(-1.645\). Since our calculated \( z \)-value \(-1.825\) is less than \(-1.645\), we reject \( H_0 \), confirming last-minute filers receive lower refunds.

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Key Concepts

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

Test Statistic

The test statistic is a vital element in hypothesis testing. It helps us determine whether to reject the null hypothesis, based on sample data. In this context, a test statistic measures how far your sample mean is from the population mean, under the null hypothesis.
To calculate it, we use the formula for the mean:\[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]where:

• \( \bar{x} \) is the sample mean (910 in our exercise).
• \( \mu \) is the hypothesized population mean (1056 for early filers).
• \( \sigma \) is the known population standard deviation (1600).
• \( n \) is the sample size (400).

By substituting these values, you calculate the z-score, which tells us how many standard deviations our sample mean is from the population mean.
In the exercise, a z-score of -1.825 was calculated, indicating the sample mean is slightly less than two standard deviations below the population mean.

P-Value

The p-value is a probability measure that helps us assess the evidence against a null hypothesis. It indicates how likely it is to observe a test statistic as extreme as the one calculated, assuming the null hypothesis is true.
A small p-value suggests that the observed data is unlikely under the null hypothesis, and thus, it may be rejected. In the exercise, the p-value was approximately 0.0340 for the calculated z-score of -1.825.
Here's how to interpret the p-value:

• If the p-value ≤ alpha level (0.05 in this case), reject the null hypothesis. This means the sample provides strong evidence against the null hypothesis.
• If the p-value > alpha level, do not reject the null hypothesis. This means there's insufficient evidence to support the alternative hypothesis.

In this scenario, the p-value of 0.0340 is less than 0.05, leading us to reject \( H_0 \), suggesting that last-minute filers indeed get lower refunds.

Critical Value Approach

The critical value approach is another way to decide on rejecting the null hypothesis in hypothesis testing. It involves comparing the test statistic to a critical value determined from a statistical distribution such as the standard normal distribution.
In a one-tailed test, you choose a critical value from the z-table corresponding to your significance level \( \alpha \). For an \( \alpha \) of 0.05, the critical z-value is approximately -1.645.
Here's the process:

• Find the critical value from a z-table using \( \alpha \).
• Compare this with your test statistic.

If the test statistic exceeds the critical value in the negative direction (e.g., is more negative), you reject the null hypothesis.
In this exercise, the test statistic of -1.825 was more negative than -1.645, meaning we reject \( H_0 \) and conclude that late filers get lower refunds.

Standard Normal Distribution

The standard normal distribution is a foundational concept in statistics. It is a normal distribution with a mean of 0 and a standard deviation of 1. It's essential in hypothesis testing for determining probabilities and critical values.
When we standardize a normal random variable, it transforms into a standard normal variable. This allows us to use z-tables to find probabilities related to our test statistic.
Some key features of the standard normal distribution include:

• Symmetrical shape around the mean (0).
• Tails that approach, but never actually touch, the horizontal axis.
• 68% of values lie within one standard deviation of the mean.
• 95% of values lie within two standard deviations.

In hypothesis testing, the standard normal distribution helps us find probabilities and critical values that determine the strength of evidence against the null hypothesis.