Problem 12 - Basic Statistics for Business and Economics with Student

Question

The management of White Industries is considering a new method of assembling its golf cart. The present method requires 42.3 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster?

Step-by-Step Solution

Answer

The new method is significantly faster.

Step 1: Formulate the Hypotheses

The hypotheses for this test are established to determine if the new method significantly reduces assembly time. The null hypothesis (H0) states that there is no reduction in time, meaning the new method takes the same time, or more, than the old one. The alternative hypothesis (H1) posits that the new method takes less time. Mathematically, these are:- Null Hypothesis (H0): \( \mu \geq 42.3 \)- Alternative Hypothesis (H1): \( \mu < 42.3 \)

Step 2: Select the Appropriate Test

Given that we have a sample mean, sample standard deviation, and wish to compare this against a known population mean, we will use a one-sample t-test for means. This is suitable because the population standard deviation is unknown, and the sample size is relatively small (n = 24).

Step 3: Calculate the Test Statistic

The test statistic for a one-sample t-test is calculated with the formula:\[t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\]where \(\bar{x} = 40.6\ \text{minutes}\), \(\mu_0 = 42.3\ \text{minutes}\), \(s = 2.7\ \text{minutes}\), and \(n = 24\). Substituting these values in:\[t = \frac{40.6 - 42.3}{\frac{2.7}{\sqrt{24}}} \approx \frac{-1.7}{0.55} \approx -3.09\]

Step 4: Determine the Critical Value

For a one-tailed test at the 0.10 level of significance and 23 degrees of freedom (n-1 = 24-1 = 23), we use a t-distribution table. The critical value of t for 0.10 significance and 23 degrees of freedom is approximately -1.319.

Step 5: Make the Decision

Compare the calculated t-statistic to the critical value. If the t-statistic is less than the critical value, we reject the null hypothesis. Here, the calculated t-statistic is \(-3.09\), which is less than \(-1.319\). Thus, we reject the null hypothesis.

Step 6: Conclusion

Since we reject the null hypothesis, there is significant evidence at the 0.10 level of significance to support the claim that the new assembly method is faster than the current method.

Key Concepts

t-test

The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. In the context of our exercise, we are specifically using a one-sample t-test.

This type of test is appropriate when you have:

• A single sample mean you want to compare against a known population mean.
• An unknown population standard deviation.
• A small sample size (usually less than 30).

For the golf cart assembly time problem, the objective of the t-test is to find out if the new method results in a mean assembly time that is significantly less than 42.3 minutes, which is the average assembly time for the current method.

Using the formula: \[t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \] where \(\bar{x}\) (sample mean) = 40.6 minutes, \(\mu_0\) (population mean) = 42.3 minutes, and \(s\) (sample standard deviation) = 2.7 minutes, we determine if the sample mean significantly differs from the population mean.

significance level

The significance level, often denoted by \(\alpha\), is the threshold for determining whether a hypothesis test's result is statistically significant. In hypothesis testing, the significance level represents the probability of rejecting the null hypothesis when it is true, also known as the Type I error rate.

In this exercise, the significance level is set at 0.10.

• This means that there is a 10% risk of concluding that the new method is faster when, in fact, it is not.
• The choice of 0.10 indicates that we're relatively tolerant of Type I errors, often because the consequences or cost of such an error are not too severe.

By setting this level, we compared our calculated t-statistic with a critical value that is derived from the t-distribution at this level of significance. If the t-statistic falls within the critical region, we reject the null hypothesis.

null hypothesis

The null hypothesis, denoted as \(H_0\), is a statement in hypothesis testing that indicates no effect or no difference. It is what we seek to disprove in order to support the alternative hypothesis.

For the problem at hand:

• The null hypothesis is that the mean assembly time using the new method is greater than or equal to the time of the current method, which is 42.3 minutes.
• Mathematically, this is expressed as \( \mu \geq 42.3 \) minutes.

A critical part of conducting hypothesis testing is to assume that the null hypothesis is true until proven otherwise.

In our scenario, rejecting \(H_0\) would mean providing enough statistical evidence that the new assembly method is indeed faster. If our calculations show that the observed data is unlikely under this null hypothesis, we can comfortably reject it.

alternative hypothesis

The alternative hypothesis, represented as \(H_1\) or \(H_a\), is a statement that suggests the presence of an effect or a difference. It is essentially the inverse of the null hypothesis and what you aim to support through your data.

In the context of the exercise:

• The alternative hypothesis claims that the mean assembly time using the new method is less than 42.3 minutes.
• This is symbolized by \( \mu < 42.3 \) minutes.

Establishing this hypothesis allows researchers to determine whether the new method indeed offers a statistically significant improvement over the old method.

When the null hypothesis is rejected in favor of \(H_1\), it provides evidence that the new method might lead to faster assembly times, a desirable outcome that supports changes in practice based on the data.