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91Ó°ÊÓ

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?

Short Answer

Expert verified
Yes, the new method is statistically faster at the 0.10 significance level.

Step by step solution

01

State the Hypotheses

We need to establish our null and alternative hypotheses. The null hypothesis is that the mean assembly time is equal to 42.3 minutes, as with the current method. The alternative hypothesis is that the mean assembly time is less than 42.3 minutes. Formally, we write:\[ H_0: \mu = 42.3 \]\[ H_a: \mu < 42.3 \]
02

Choose a Significance Level

We are given the significance level \( \alpha = 0.10\). This is our threshold for deciding whether to reject the null hypothesis.
03

Compute the Test Statistic

We will use the t-test statistic because the population standard deviation is unknown and the sample size is small (n = 24). The formula for the t-test statistic is:\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]where \( \bar{x} = 40.6 \), \( \mu_0 = 42.3 \), \( s = 2.7 \), and \( n = 24 \). Substitute these values into the formula:\[ t = \frac{40.6 - 42.3}{2.7/\sqrt{24}} \approx -3.166 \]
04

Determine the Critical Value

For a one-tailed test at \( \alpha = 0.10 \) with \( n - 1 = 23 \) degrees of freedom, we look up the t-distribution table or use technology to find the critical t-value. The critical t-value is approximately -1.319.
05

Make a Decision

Compare the calculated t-statistic with the critical t-value. Our calculated t-statistic is -3.166. Since -3.166 is less than -1.319, we fall into the rejection region.
06

Conclusion

Since our t-statistic is in the rejection region, we reject the null hypothesis. There is sufficient evidence at the \( 0.10 \) significance level to conclude that the assembly time using the new method is faster.

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Key Concepts

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

t-test
When dealing with small sample sizes or unknown population standard deviations, the t-test is a powerful statistical tool. It helps us determine if there is a significant difference between the means of our sample and the population.
This is particularly useful in scenarios where we want to check if a new method or treatment is effective.

In our example, we are interested in testing if the new method of assembling golf carts reduces the assembly time. Using the t-test allows us to analyze the sample data, compare it against a known average, and make data-driven decisions.
Here’s how it works:
  • Calculate the test statistic using the formula: \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \] where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.
  • This gives us a "t" value that indicates how much our sample's mean differs from the hypothesized population mean.
  • We then compare this "t" value against the critical value to determine if our result is significant.
The result provides us insight into whether we can confidently say that the observed changes are meaningful and not just due to random variance.
null hypothesis
The null hypothesis, often denoted as \( H_0 \), is a statement used in statistical hypothesis testing that assumes there is no effect or difference. It represents the status quo or baseline expectation.
In hypothesis testing, our goal is usually to disprove or reject this assumption based on the data we collect.

Applying this to our example:
  • The null hypothesis is that the new golf cart assembly method does not result in any change, meaning the mean assembly time is still 42.3 minutes.
  • We express this formally as \( H_0: \mu = 42.3 \).
The null hypothesis is central to statistical testing because it is the theory we aim to challenge with evidence from our data.
If our test results show a significant effect, we may reject the null hypothesis, suggesting that change or difference is indeed present.
significance level
The significance level, represented by \( \alpha \), is a threshold determined before conducting a hypothesis test. It helps control how willing we are to make an error when rejecting the null hypothesis.
Common significance levels are 0.05, 0.10, and 0.01, each corresponding to different levels of risk acceptance.

In the context of our example:
  • We use a significance level of \( \alpha = 0.10 \), indicating we are willing to accept a 10% chance of incorrectly rejecting the null hypothesis.
  • This level helps balance the risk of Type I error (falsely rejecting a true null hypothesis) against the need to detect an effect if it actually exists.
Choosing an adequate significance level is crucial. It impacts our decision-making process, helping us to remain systematic and evidence-based.
When set too low, it could prevent us from detecting meaningful differences. Conversely, a high \( \alpha \) increases the chance of a false positive.
critical value
The critical value is an important threshold in hypothesis testing. It is the point against which we compare our test statistic to decide whether to reject the null hypothesis.
Critical values vary based on the significance level and the type of test being conducted (one-tailed or two-tailed).

For our assembly time example:
  • The critical value corresponds to the t-distribution value for a one-tailed test at \( \alpha = 0.10 \) with 23 degrees of freedom.
  • We found this critical value to be approximately -1.319 from the t-table or statistical software.
If our calculated t-statistic falls beyond this critical value, we enter the rejection region. In our scenario, this means rejecting the null hypothesis because we have enough evidence indicating that assembly time with the new method is faster.
The critical value is central as it provides the cutoff for determining whether observed data are statistically significant.

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Most popular questions from this chapter

The Rutter Nursery Company packages their pine bark mulch in 50 -pound bags. From a long history, the production department reports that the distribution of the bag weights follows the normal distribution and the standard deviation of this process is 3 pounds per bag. At the end of each day, Jeff Rutter, the production manager, weighs 10 bags and computes the mean weight of the sample. Below are the weights of 10 bags from today's production. $$\begin{array}{|lllllllll|}\hline 45.6 & 47.7 & 47.6 & 46.3 & 46.2 & 47.4 & 49.2 & 55.8 & 47.5 & 48.5 \\\\\hline\end{array}$$ a. Can Mr. Rutter conclude that the mean weight of the bags is less than 50 pounds? Use the .01 significance level. b. In a brief report, tell why Mr. Rutter can use the \(z\) distribution as the test statistic. c. Compute the \(p\) -value.

The MacBurger restaurant chain claims that the waiting time of customers for service is normally distributed, with a mean of 3 minutes and a standard deviation of 1 minute. The quality-assurance department found in a sample of 50 customers at the Warren Road MacBurger that the mean waiting time was 2.75 minutes. At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes?

The following hypotheses are given. $$\begin{array}{l}H_{0}: \pi=.40 \\\H_{1}: \pi \neq .40\end{array}$$ A sample of 120 observations revealed that \(p=.30 .\) At the .05 significance level, can the null hypothesis be rejected? a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

A recent article in USA Today reported that a job awaits only one in three new college graduates. The major reasons given were an overabundance of college graduates and a weak economy. A survey of 200 recent graduates from your school revealed that 80 students had jobs. At the .02 significance level, can we conclude that a larger proportion of students at your school have jobs?

The following information is available. $$\begin{array}{l}H_{0}: \mu \leq 10 \\\H_{1}: \mu>10\end{array}$$ The sample mean is 12 for a sample of \(36 .\) The population follows the normal distribution and the standard deviation is \(3 .\) Use the .02 significance level.

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