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A medical laboratory claims that the mean turn-around time for performance of a battery of tests on blood samples is 1.88 business days. The manager of a large medical practice believes that the actual mean is larger. A random sample of 45 blood samples yielded mean 2.09 and sample standard deviation 0.13 day. Perform the relevant test at the \(10 \%\) level of significance, using these data.

Short Answer

Expert verified
The test supports that the mean turn-around time is greater than 1.88 days at the 10% significance level.

Step by step solution

01

Define Hypotheses

We need to determine whether the actual mean turn-around time is larger than the claimed 1.88 days. Therefore, we set up our hypotheses as follows: \( H_0: \mu = 1.88 \) days (the null hypothesis) and \( H_a: \mu > 1.88 \) days (the alternative hypothesis).
02

Determine Test Statistic

Since the sample size is 45, which is greater than 30, we use the z-test for our test statistic. The formula for the z-test statistic is \( z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \), where \( \bar{x} = 2.09 \), \( \mu = 1.88 \), \( s = 0.13 \), and \( n = 45 \).
03

Calculate Test Statistic

Plug in the values into the formula: \( z = \frac{2.09 - 1.88}{\frac{0.13}{\sqrt{45}}} \). First, calculate the denominator: \( \frac{0.13}{\sqrt{45}} \approx 0.0194 \). Therefore, \( z = \frac{0.21}{0.0194} \approx 10.82 \).
04

Determine Critical Value

For a right-tailed test at the 10% level of significance, we find the critical z-value from the z-table. The critical z-value for \( \alpha = 0.10 \) is approximately 1.28.
05

Compare Test Statistic and Critical Value

We compare the calculated z-value (10.82) with the critical z-value (1.28). Since 10.82 is much greater than 1.28, we reject the null hypothesis.
06

Conclusion

Since we rejected the null hypothesis, we conclude that there is sufficient evidence at the 10% significance level to support the manager's claim that the actual mean turn-around time is greater than 1.88 days.

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Key Concepts

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

Understanding the Z-Test
The z-test is a statistical method used for hypothesis testing. It helps you decide if there is enough evidence to reject a null hypothesis. The test is especially useful when dealing with large sample sizes, typically greater than 30.

In a z-test, you compare the sample mean to a known value (like the population mean). You calculate the z-test statistic to see how the sample differs from what is expected if the null hypothesis were true. This test offers an effective way to judge the probability that an observed difference could have happened randomly.

The z-test formula is:
  • \( z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \)
Here, \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size. Using these inputs, the formula provides a z-value, which helps you determine how far your sample mean is from the hypothesized population mean, expressed in standard deviation units.
Exploring the Null Hypothesis
The null hypothesis (denoted as \( H_0 \)) represents a default position that there is no effect or no difference. In the context of hypothesis testing, it proposes that any kind of difference or significance you see in a set of data is due to chance. It's the cornerstone of hypothesis testing since the main goal is to know whether to reject this baseline notion.

For example, in the exercise, the null hypothesis was \( H_0: \mu = 1.88 \). This hypothesis states that the mean turn-around time is exactly 1.88 days. When you conduct the test, you are trying to gather enough evidence to prove this hypothesis false or unable to be rejected.

In practice, if your test statistic indicates that the null hypothesis is likely not true, you may reject \( H_0 \) with some level of confidence. This allows for meaningful conclusions based on statistical evidence.
Clarifying the Alternative Hypothesis
The alternative hypothesis (denoted as \( H_a \)) is your test or research hypothesis. It represents a statement that is believed to be true and one that you aim to support through evidence from the data. Unlike the null hypothesis, the alternative claims that there is an effect, a difference, or a change.

In the exercise, the alternative hypothesis was \( H_a: \mu > 1.88 \). This reflects that the mean turn-around time is greater than what was initially claimed.

When testing hypotheses, the goal is to see if there is enough statistical evidence to reject the null hypothesis in favor of the alternative hypothesis. The choice between the null and alternative depends on the test statistics and the evidence gathered during analysis. By establishing a clear alternative hypothesis, you focus your investigation on the possibility of an effect or difference being present.
Understanding Significance Level
The significance level, denoted by \( \alpha \), is a critical value that determines the threshold for rejecting the null hypothesis. It represents the probability of committing a Type I error, which occurs when you wrongly reject a true null hypothesis.

A common choice for \( \alpha \) is 0.05, but it can vary depending on the context of the study. In our exercise, the significance level was set at 10% (0.10). This means you are willing to accept a 10% risk of rejecting \( H_0 \) when it is actually true.

When you evaluate the test statistic, if it is beyond the critical value corresponding to your significance level, you reject the null hypothesis. Hence, the significance level is a measure of your confidence in the test results. Picking the right \( \alpha \) helps balance the risk of errors against the need for statistical certainty. In a right-tailed test, as in our case, the calculated critical z-value reflects this threshold needed to confirm statistically meaningful results.

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Most popular questions from this chapter

A grocery store chain has as one standard of service that the mean time customers wait in line to begin checking out not exceed 2 minutes. To verify the performance of a store the company measures the waiting time in 30 instances, obtaining mean time 2.17 minutes with standard deviation 0.46 minute. Use these data to test the null hypothesis that the mean waiting time is 2 minutes versus the alternative that it exceeds 2 minutes, at the \(10 \%\) level of significance.

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A lawyer believes that a certain judge imposes prison sentences for property crimes that are longer than the state average 11.7 months. He randomly selects 36 of the judge's sentences and obtains mean 13.8 and standard deviation 3.9 months. a. Perform the test at the \(1 \%\) level of significance using the critical value approach. b. Compute the observed significance of the test. c. Perform the test at the \(1 \%\) level of significance using the \(p\) -value approach. You need not repeat the first three steps, already done in part (a).

An insurance company states that it settles \(85 \%\) of all life insurance claims within 30 days. A consumer group asks the state insurance commission to investigate. In a sample of 250 life insurance claims, 203 were settled within 30 days. a. Test whether the true proportion of all life insurance claims made to this company that are settled within 30 days is less than \(85 \%,\) at the \(5 \%\) level of significance. b. Compute the observed significance of the test.

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