/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 84 Dartmouth Distribution Warehouse... [FREE SOLUTION] | 91Ó°ÊÓ

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Dartmouth Distribution Warehouse makes deliveries of a large number of products to its customers. To keep its customers happy and satisfied, the company's policy is to deliver on time at least \(90 \%\) of all the orders it receives from its customers. The quality control inspector at the company quite often takes samples of orders delivered and checks to see whether this policy is maintained. A recent sample of 90 orders taken by this inspector showed that 75 of them were delivered on time. a. Using a \(2 \%\) significance level, can you conclude that the company's policy is maintained? b. What will your decision be in part a if the probability of making a Type I error is zero? Explain.

Short Answer

Expert verified
a. No, at a \(2 \%\) level of significance, we do not have enough evidence to conclude that the company's policy is not being maintained. b. If the Type I error probability is zero, the decision will be the same, not to reject the null hypothesis.

Step by step solution

01

Formulate the Hypotheses

First, formulate the null hypothesis (H0) and the alternative hypothesis (Ha). Here, H0 would be that the warehouse makes on time deliveries at least \(90 \%\) of the time, or \(p = 0.9\), and Ha would be that the rate of on-time deliveries is less than \(90 \%\) , or \(p < 0.9\).
02

Calculate Sample Proportion

The sample proportion on time deliveries (\(p_s\)) is the number of on time deliveries (\(x\)) divided by total number of orders sampled (\(n\)). In this case, \(p_s = \frac{75}{90} = 0.8333\).
03

Perform the Z-Test

The z score measures the number of standard deviations an element is from the mean. In this case, calculate z score using the formula \(z = \frac{p_s - p}{\sqrt{\frac{p(1-p)}{n}}}\), where \(p\) is the proportion of on time deliveries according to warehouse policy, \(p_s\) is the sample proportion, and \(n\) is the number of orders sampled. Substituting the values, the test statistic is \(z = \frac{0.8333 - 0.9}{\sqrt{\frac{0.9 * (1 - 0.9)}{90}}} = -1.66\)
04

Find the Critical Value and Make Decision

The critical value for a significance level (alpha) of \(2 \%\) (or 0.02) in a one-tailed test (which is the case here) can be obtained from a z-table or z-calculator, and it is roughly -2.05. Since our calculated z score is greater than the critical value (-1.66 > -2.05), we fail to reject H0. Therefore, there isn't enough evidence at \(2 \%\) significance level to conclude that the warehouse's on-time delivery policy isn't being maintained.
05

Considering Type I error

A Type I error happens when the null hypothesis is true but is rejected. It is specified by the level of significance. In this case, if the probability of making a Type I error is zero, it means the level of significance is also zero. This would mean an extreme level of caution in rejecting the null hypothesis. Even if there was very strong evidence against the null hypothesis, it would still not be rejected, as that would risk making a Type I error. So, the decision would still be to not reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is essentially a statement that there is no effect or no difference, and it serves as the starting point for statistical testing. In the context of the Dartmouth Distribution Warehouse, the null hypothesis (denoted as \(H_0\)) claims that on-time deliveries occur at least \(90\%\) of the time. This implies that the system operates as expected and there is no evidence of underperformance.
The null hypothesis is what we actually test against, and our goal is to see if we have sufficient evidence to reject it. It's like assuming "status quo" unless proven otherwise.
Keeping this hypothesis in mind helps not only protect against false positives but also establishes a baseline for all statistical evaluations.
Alternative Hypothesis
Opposite to the null hypothesis, the alternative hypothesis presents a statement that is to be considered if the null hypothesis is deemed unlikely. In our warehouse example, the alternative hypothesis (denoted as \(H_a\)) proposes that the proportion of on-time deliveries is less than \(90\%\).
This hypothesis embodies the notion of "change" or "difference" from what is commonly accepted. It's what we aim to support through the evidence collected. In hypothesis testing, proving the alternative hypothesis means we have found sufficient evidence to indicate a deviation from the status quo, hence calling the original assumption in question.
Z-Test
The z-test is a type of statistical test that helps determine whether there is a significant difference between sample data and a population parameter. Specifically, it measures how far away the sample data is, in terms of standard deviations, from the assumed population mean.
In the warehouse scenario, the z-test is employed to ascertain the difference between the sample proportion of on-time deliveries (\(0.8333\)) and the stated policy of \(90\%\). The test statistic calculated as \(z = \frac{p_s - p}{\sqrt{\frac{p(1-p)}{n}}}\), where \(p\) and \(p_s\) are the known and sample proportions respectively, determines whether the deviation is statistically significant.
The interpretation hinges on comparing the observed z-value to a critical z-value which was identified to be \(-2.05\) for a \(2\%\) significance level. Finding the z-test essential allows for effective decision-making regarding the null hypothesis.
Significance Level
The significance level, typically denoted by \(\alpha\), represents the probability of rejecting the null hypothesis when it is actually true. It acts as a threshold for determining whether a test result is statistically significant.
In the provided example, a \(2\%\) significance level implies there is only a \(2\%\) risk of incorrectly rejecting the null hypothesis (known as a Type I error). This low significance level indicates a stringent criterion for evidence needed to claim a deviation from the warehouse's delivery policy.
Setting a significance level is crucial as it directly affects the outcome of hypothesis testing. A low \(\alpha\) means more stringent criteria, reducing the likelihood of considering a change when there is none, ensuring decisions are made with utmost care and precision.

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Most popular questions from this chapter

Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of the rods are approximately normally distributed and vary slightly. It is known that when the machine is working properly, the mean length of the rods is 36 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to \(.035\) inch. The quality control department at the company takes a sample of 20 such rods every week, calculates the mean length of these rods, and tests the null hypothesis, \(\mu=36\) inches, against the alternative hypothesis, \(\mu \neq 36\) inches. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 20 rods produced a mean length of \(36.015\) inches. a. Calculate the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, will the quality control inspector decide to stop the machine and adjust it if he chooses the maximum probability of a Type I error to be .02? What if the maximum probability of a Type I error is \(10 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.02\). Does the machine need to be adjusted? What if \(\alpha=.10 ?\)

A mail-order company claims that at least \(60 \%\) of all orders are mailed within 48 hours. From time to time the quality control department at the company checks if this promise is fulfilled. Recently the quality control department at this company took a sample of 400 orders and found that 208 of them were mailed within 48 hours of the placement of the orders.

Consider the following null and alternative hypotheses: $$ H_{0}: p=.82 \text { versus } H_{1}: p \neq .82 $$ A random sample of 600 observations taken from this population produced a sample proportion of \(.86\). a. If this test is made at a \(2 \%\) significance level, would you reject the null hypothesis? Use the critical-value approach. b. What is the probability of making a Type I error in part a? c. Calculate the \(p\) -value for the test. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.025\) ? What if \(\alpha=.01 ?\)

Consider the following null and alternative hypotheses: $$ H_{0}=p=.44 \text { versus } H_{1}: p<.44 $$ A random sample of 450 observations taken from this population produced a sample proportion of \(.39 .\) a. If this test is made at a \(2 \%\) significance level, would you reject the null hypothesis? Use the critical-value approach. b. What is the probability of making a Type I error in part a? c. Calculate the \(p\) -value for the test. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.01 ?\) What if \(\alpha=.025\) ?

A past study claimed that adults in America spent an average of 18 hours a week on leisure activities. A researcher wanted to test this claim. She took a sample of 12 adults and asked them about the time they spend per week on leisure activities. Their responses (in hours) are as follows. \(\begin{array}{lllllllllllll}13.6 & 14.0 & 24.5 & 24.6 & 22.9 & 37.7 & 14.6 & 14.5 & 21.5 & 21.0 & 17.8 & 21.4\end{array}\) Assume that the times spent on leisure activities by all American adults are normally distributed. Using a \(10 \%\) significance level, can you conclude that the average amount of time spent by American adults on leisure activities has changed? (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections 3.1.1 and \(3.2 .2\) of Chapter \(3 .\) Then make the test of hypothesis about \(\mu .\) )

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