Problem 52 - Basic Statistics for Business and Economics with Student

Question

From past experience a television manufacturer found that 10 percent or less of its sets needed repair in the first two years of operation. In a sample of 50 sets manufactured two years ago, 9 needed repair. At the .05 significance level, has the percent of sets needing repair increased? Determine the \(p\) -value.

Step-by-Step Solution

Answer

Yes, the percent of sets needing repair has increased; the p-value is approximately 0.029.

Step 1: Define Hypotheses

Formulate the null and alternative hypotheses for the proportion test. The null hypothesis \( H_0 \) is that the proportion of television sets needing repair is 10% or less, \( p \leq 0.10 \). The alternative hypothesis \( H_a \) is that the proportion is greater than 10%, \( p > 0.10 \).

Step 2: State the Significance Level

The problem states a significance level of \( \alpha = 0.05 \). This will be used to determine the rejection criteria for the null hypothesis.

Step 3: Calculate the Sample Proportion

Calculate the sample proportion \( \hat{p} \) of TV sets that needed repair. This is done by dividing the number of sets that needed repair by the total number of sets in the sample: \( \hat{p} = \frac{9}{50} = 0.18 \).

Step 4: Calculate the Test Statistic

Use the formula for the test statistic for a proportion: \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}} \), where \( p_0 = 0.10 \) and \( n = 50 \). Substitute the values to get \( z = \frac{0.18 - 0.10}{\sqrt{\frac{0.10 \times 0.90}{50}}} \).

Step 5: Compute Numerical Result for Test Statistic

Calculate the numerical result for the test statistic: \( z = \frac{0.08}{\sqrt{0.0018}} \approx \frac{0.08}{0.0424} \approx 1.89 \).

Step 6: Determine the Critical Value and Decision Rule

For a right-tailed test at \( \alpha = 0.05 \), the critical value of \( z \) is approximately 1.645. If the calculated \( z \)-value is greater than 1.645, we reject the null hypothesis.

Step 7: Make a Decision

Since the computed \( z \)-value (1.89) is greater than the critical value (1.645), we reject the null hypothesis. There is significant evidence at the 0.05 significance level to conclude that the proportion of TV sets needing repair has increased.

Step 8: Calculate the p-value

The \( p \)-value corresponds to the calculated \( z \)-value in a standard normal distribution. Using a z-table or calculator, find the area to the right of \( z = 1.89 \); this gives a \( p \)-value of approximately 0.029.

Key Concepts

Understanding Proportion Test

A proportion test is a statistical tool used when you want to compare a sample proportion to a known proportion or to compare two sample proportions. In this exercise, we examine whether the proportion of television sets needing repair exceeds the previous experience rate of 10%.

The proportion test aims to determine if the observed sample proportion, in this case, 18% of the TV sets needing repair, indicates a significant change from the expected 10%. This involves setting up a null hypothesis ( ull hypothesis ) and an alternative hypothesis ( alternate hypothesis ):

• Null Hypothesis ( ull hypothesis ): The proportion of sets needing repair is 10% or less.
• Alternate Hypothesis ( alternate hypothesis ): The proportion of sets needing repair is greater than 10%.

Based on the outcome of the test statistic and the critical value, we decide whether there is enough evidence to reject the null hypothesis.

Significance Level Demystified

The significance level, denoted by \( \alpha \), is a threshold used to decide whether to reject the null hypothesis. In this exercise, the significance level is set at 0.05. This signifies a 5% risk of concluding that the proportion of sets needing repair has increased when it hasn't.

Choosing a significance level involves a tradeoff:

• A lower \( \alpha \) decreases the chances of a Type I error (incorrectly rejecting the null hypothesis), but may increase the chance of a Type II error (failing to detect an actual effect).
• A higher \( \alpha \) increases the risk of a Type I error but makes it easier to detect an actual effect.

The significance level is especially crucial in hypothesis testing, guiding the researcher in making informed decisions about rejecting or not rejecting the null hypothesis.

P-Value Calculation Explained

The \( p \)-value in hypothesis testing helps determine the strength of the results. It's a measure of the probability of observing a test statistic as extreme as, or more extreme than, the observed value, assuming the null hypothesis is true.

In our case, the calculated \( z \)-value was 1.89. Using a standard normal distribution (Z-table), you find the area to the right of this \( z \)-value to get the \( p \)-value.

• The smaller the \( p \)-value, the stronger the evidence against the null hypothesis.
• If the \( p \)-value is less than the significance level (\( \alpha = 0.05 \)), we reject the null hypothesis.
• In this exercise, we found a \( p \)-value of approximately 0.029, which is less than our significance level of 0.05, supporting the rejection of the null hypothesis.

The \( p \)-value provides a quantitative measure for decision making, highlighting whether the evidence is strong enough to make a conclusion about the population proportion.

Z-Test Unveiled

A Z-test is a type of hypothesis test used for situations where the data can be described using statistics such as a mean. For proportions, like in the exercise, we use the Z-test to compute whether the observed sample proportion significantly differs from a specified value.

In this scenario, the Z-test formula for a sample proportion is: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}} \]

• \( \hat{p} \) is the sample proportion (0.18).
• \( p_0 \) is the hypothesis proportion (0.10).
• \( n \) is the sample size (50).

Plugging in these values gives a \( z \)-value of 1.89. This value is compared to the critical \( z \)-value to determine statistical significance:

• If the calculated \( z \)-value is greater than the critical \( z \)-value (for \( \alpha = 0.05 \), critical \( z \)-value is about 1.645 for a right-tailed test), the null hypothesis is rejected.

Using the Z-test, we concluded that there is substantial evidence that the proportion of television sets needing repair has increased beyond the original expectation.