Problem 49 - Basic Statistics for Business and Economics with Student

Question

Tina Dennis is the comptroller for Meek Industries. She believes that the current cash-flow problem at Meek is due to the slow collection of accounts receivable. She believes that more than 60 percent of the accounts are in arrears more than three months. A random sample of 200 accounts showed that 140 were more than three months old. At the .01 significance level, can she conclude that more than 60 percent of the accounts are in arrears for more than three months? \(?\)

Step-by-Step Solution

Answer

Yes, Tina can conclude that more than 60% of accounts are in arrears for more than three months.

Step 1: Define Hypotheses

First, we need to set up our null and alternate hypotheses. Let \( p \) be the proportion of accounts in arrears more than three months. The null hypothesis \( H_0 \) states that \( p \leq 0.6 \) (60%). The alternative hypothesis \( H_a \) is that \( p > 0.6 \). We will test these hypotheses using a significance level of \( \alpha = 0.01 \).

Step 2: Determine Sample Proportion

Calculate the sample proportion \( \hat{p} \). Given the sample size \( n = 200 \) and the number of successful accounts \( x = 140 \), the sample proportion is calculated as follows: \( \hat{p} = \frac{x}{n} = \frac{140}{200} = 0.7 \).

Step 3: Calculate Test Statistic

We will use a z-test for proportions to calculate the test statistic. The test statistic formula is \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \), where \( p_0 = 0.6 \). Substituting the values, we have \[ z = \frac{0.7 - 0.6}{\sqrt{\frac{0.6 \times 0.4}{200}}} \approx \frac{0.1}{0.0346} \approx 2.89 \].

Step 4: Determine Critical Value

At the \( \alpha = 0.01 \) significance level and for a one-tailed test, we find the critical z-value from the standard normal distribution table, which is approximately 2.33. This means if our test statistic is greater than 2.33, we can reject the null hypothesis.

Step 5: Make a Decision

Compare the calculated test statistic to the critical value. Our test statistic \( z = 2.89 \) is greater than the critical value of 2.33. Therefore, we reject the null hypothesis.

Key Concepts

Proportion Test

A proportion test is a statistical procedure used to determine if a sample proportion is significantly different from a hypothesized population proportion. This type of test is particularly useful when dealing with categorical data, where the outcome is typically binomial (e.g., success/failure, yes/no). Here, the focus is on whether the proportion of a specific outcome in a sample is representative of the stated proportion in the broader population.

In the context of Tina Dennis's problem, we are interested in whether more than 60% of the accounts at Meek Industries are late by more than three months. The proportion test helps us objectively decide this, using the sample data and statistical tools.

• Sample Size: The number of observations or accounts Tina analyzed, which is 200.
• Sample Proportion (\( \hat{p} \)): The observed proportion in the sample, calculated as 0.7 (140 out of 200 accounts).
• Hypothesized Population Proportion (\( p_0 \)): The value Tina is testing against, 0.6 or 60%.

The proportion test, using these values, helps determine if the observed sample proportion of 0.7 differs significantly from the hypothesized proportion of 0.6.

Significance Level

The significance level, often denoted by \( \alpha \), is a critical concept in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true, essentially measuring the potential for a Type I error. In most hypothesis testing scenarios, common significance levels are 0.05, 0.01, or 0.10.

In Tina's scenario, the significance level chosen is \( \alpha = 0.01 \). This means there is a 1% risk of concluding that more than 60% of accounts are in arrears when in fact they are not. A lower significance level such as 0.01 implies a stricter criterion for rejecting the null hypothesis.

By setting the significance level to 0.01, Tina expresses a strong desire for confidence in her conclusions, minimizing the possibility of mistaken declarations about the accounts' arrears status. Thus, only if the evidence from the sample is very persuasive, reflected by a sufficiently high test statistic, will the null hypothesis be rejected at this rigorous significance level.

Null Hypothesis

The null hypothesis, noted as \( H_0 \), is a fundamental part of hypothesis testing, serving as the default claim to be tested. It usually reflects the "status quo" or a position of no effect or no difference. The objective in hypothesis testing is to challenge this null hypothesis with evidence from the collected data.

For Meek Industries' case, Tina's null hypothesis is \( H_0: p \leq 0.6 \). This suggests that Tina initially assumes 60% or less of the accounts are more than three months overdue. This hypothesis serves as a baseline to test against the collected data.

The null hypothesis stands unless sufficient statistical evidence advocates for its rejection. In statistical terms, we use the sample data to see if it provides enough proof against the null hypothesis under the pre-set significance level.

Alternative Hypothesis

The alternative hypothesis, denoted as \( H_a \), represents a statement contradictory to the null hypothesis. It indicates what you aim to provide evidence for through the test. In hypothesis testing, you look for data evidence to reject the null hypothesis in favor of the alternative hypothesis.

For the scenario Tina is analyzing, the alternative hypothesis is \( H_a: p > 0.6 \). This suggests that more than 60% of the accounts are overdue by more than three months, contrary to what the null hypothesis asserts.

In this analysis, the alternative hypothesis is one-sided, as Tina is specifically interested in finding evidence that supports the idea of a greater proportion than the stated 60%. The outcome of the hypothesis test depends on whether the evidence (z-score) significantly surpasses the critical value at the established significance level, thus justifying the rejection of the null hypothesis and accepting the alternative.