/*! 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 38 A random sample of 150 recent do... [FREE SOLUTION] | 91Ó°ÊÓ

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A random sample of 150 recent donations at a blood bank reveals that 82 were type A blood. Does this suggest that the actual percentage of type A donations differs from \(40 \%\), the percentage of the population having type A blood? Carry out a test of the appropriate hypotheses using a significance level of .01. Would your conclusion have been different if a significance level of \(.05\) had been used?

Short Answer

Expert verified
At \(\alpha = 0.01\), do not reject the null hypothesis as \(z\) is within critical limits. At \(\alpha = 0.05\), still do not reject it.

Step by step solution

01

Define the Hypotheses

We want to test if the percentage of type A blood donations in the sample differs from the population percentage. Define the null hypothesis \(H_0: p = 0.40\), where \(p\) is the proportion of type A donations. The alternative hypothesis is \(H_a: p eq 0.40\). This is a two-tailed test.
02

Calculate the Test Statistic

First, calculate the sample proportion \(\hat{p} = \frac{82}{150}\). Next, use the formula for the standard error \(SE = \sqrt{\frac{p(1-p)}{n}}\) where \(n\) is the sample size. Then calculate the test statistic \(z = \frac{\hat{p} - p}{SE}\).
03

Determine the Critical Value and Compare

For a significance level of \(\alpha = 0.01\), the critical z-values are \(-2.576\) and \(+2.576\) for a two-tailed test. If the calculated \(z\)-value is less than \(-2.576\) or greater than \(+2.576\), reject the null hypothesis.
04

Conclusion at \(\alpha = 0.01\)

Calculate the \(z\)-value from Step 2, then compare it to the critical values from Step 3. If the \(z\)-value falls outside \(-2.576\) and \(+2.576\), conclude that there is significant evidence to suggest the actual percentage differs from \(40\%\). Otherwise, do not reject the null hypothesis.
05

Conclusion at \(\alpha = 0.05\)

The critical values for \(\alpha = 0.05\) are \(-1.96\) and \(+1.96\). Reevaluate the calculated \(z\)-value to see if it falls beyond these critical values. The null hypothesis should be rejected if \(\hat{z}\) exceeds these bounds.

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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 acts as a starting position which states that no effect or no difference is expected. It is a statement we aim to test and potentially reject. The null hypothesis is denoted by \(H_0\). In the given problem, the question involves checking whether the proportion of type A blood donations aligns with the known population proportion of 40%. Therefore, the null hypothesis is formulated as \(H_0: p = 0.40\), where \(p\) represents the proportion of type A blood donations. The purpose of the null hypothesis is to examine the evidence against it, allowing us to either accept it or reject it based on the data analysis.
Alternative Hypothesis
Contrary to the null hypothesis, the alternative hypothesis represents the statement we consider to be true if evidence suggests the null hypothesis is incorrect. It is denoted by \(H_a\) and usually signifies the presence of an effect or a difference.In our example, we are testing if the real percentage of type A blood donations deviates from 40%. This requires forming an alternative hypothesis: \(H_a: p eq 0.40\). This implies we suspect that the proportion of type A blood donations is not equal to 40%. This type of hypothesis is often what researchers aim to support through their experiments. If the data provide enough evidence, the alternative hypothesis is accepted, indicating a significant result.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold set by the researcher. It defines the probability of rejecting the null hypothesis when it is actually true, essentially controlling the risk of a false positive.In hypothesis testing, this level is crucial as it helps determine the critical values of the test statistic. For this particular exercise, a significance level of \(0.01\) is used, implying a 1% risk of erroneously rejecting the null hypothesis.Lower significance levels, like 0.01, show stringent testing, meaning the conclusions need to be supported by strong evidence. Whereas a significance level of 0.05 allows more flexibility to reject the null hypothesis if there is not overwhelming evidence against it.
Test Statistic
A test statistic is a standardized value used to decide whether to reject the null hypothesis. Based on the sampling distribution, it provides a means to compare observed data with the null hypothesis. For this problem, the test statistic is calculated using a z-test. The sample proportion \(\hat{p}\) is first determined, then the standard error \(SE\) is computed using the formula \(SE = \sqrt{\frac{p(1-p)}{n}}\), where \(n\) is the sample size. Subsequently, the z-statistic is found using the equation \(z = \frac{\hat{p} - p}{SE}\). In our exercise, this z-statistic is then compared with critical z-values to draw conclusions regarding the hypothesis. Calculating the test statistic is a pivotal step, as it quantifies how far away our sample result is from what was expected under the null hypothesis.
Critical Value
Critical values help determine the boundary for the test statistic, beyond which the null hypothesis can be rejected. They are point(s) on the scale of the test statistic, establishing regions where the null hypothesis may or may not be rejected.Given a two-tailed test with a significance level of 0.01, the critical values are found on the standard normal distribution as \(-2.576\) and \(+2.576\). This means if the calculated z-value exceeds these boundaries, the null hypothesis is rejected as it suggests strong evidence for a difference.Adjustments to the significance level, such as using \(0.05\) instead, alter the critical values, leading to different decision boundaries (\(-1.96\) and \(+1.96\)), thereby affecting hypothesis test conclusions.

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Most popular questions from this chapter

A new design for the braking system on a certain type of car has been proposed. For the current system, the true average braking distance at 40 mph under specified conditions is known to be \(120 \mathrm{ft}\). It is proposed that the new design be implemented only if sample data strongly indicates a reduction in true average braking distance for the new design. a. Define the parameter of interest and state the relevant hypotheses. b. Suppose braking distance for the new system is normally distributed with \(\sigma=10\). Let \(\bar{X}\) denote the sample average braking distance for a random sample of 36 observations. Which of the following rejection regions is appropriate: \(R_{1}=\\{\bar{x}: \bar{x} \geq 124.80\\}, R_{2}=\) \(\\{\bar{x}: \bar{x} \leq 115.20\\}, R_{3}=\\{\bar{x}:\) either \(\bar{x} \geq 125.13\) or \(\bar{x} \leq 114.87\\} ?\) c. What is the significance level for the appropriate region of part (b)? How would you change the region to obtain a test with \(\alpha=.001\) ? d. What is the probability that the new design is not implemented when its true average braking distance is actually \(115 \mathrm{ft}\) and the appropriate region from part (b) is used? e. Let \(Z=(\bar{X}-120) /(\sigma / \sqrt{n})\). What is the significance level for the rejection region \(\\{z\) : \(z \leq-2.33\\}\) ? For the region \(\\{z: z \leq-2.88\\}\) ?

A sample of 50 lenses used in eyeglasses yields a sample mean thickness of \(3.05 \mathrm{~mm}\) and a sample standard deviation of \(.34 \mathrm{~mm}\). The desired true average thickness of such lenses is \(3.20 \mathrm{~mm}\). Does the data strongly suggest that the true average thickness of such lenses is something other than what is desired? Test using \(\alpha=.05\).

Reconsider the paint-drying problem discussed in Example 9.2. The hypotheses were \(H_{0}: \mu=75\) versus \(H_{\mathrm{a}}: \mu<75\), with \(\sigma\) assumed to have value 9.0. Consider the alternative value \(\mu=74\), which in the context of the problem would presumably not be a practically significant departure from \(H_{0}\). a. For a level .01 test, compute \(\beta\) at this alternative for sample sizes \(n=100,900\), and 2500 . b. If the observed value of \(\bar{X}\) is \(\bar{x}=74\), what can you say about the resulting \(P\)-value when \(n=2500 ?\) Is the data statistically significant at any of the standard values of \(\alpha\) ? c. Would you really want to use a sample size of 2500 along with a level \(.01\) test (disregarding the cost of such an experiment)? Explain.

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge-water temperature above \(150^{\circ}, 50\) water samples will be taken at randomly selected times, and the temperature of each sample recorded. The resulting data will be used to test the hypotheses \(H_{0}: \mu=150^{\circ}\) versus \(H_{\mathrm{a}}: \mu>150^{\circ}\). In the context of this situation, describe type I and type II errors. Which type of error would you consider more serious? Explain.

The recommended daily dietary allowance for zinc among males older than age 50 years is \(15 \mathrm{mg} /\) day. The article "Nutrient Intakes and Dietary Pattems of Older Americans: A National Study" (J. Gerontol., 1992: M145-150) reports the following summary data on intake for a sample of males age 65-74 years: \(n=115, \bar{x}=11.3\), and \(s=6.43\). Does this data indicate that average daily zinc intake in the population of all males age 65-74 falls below the recommended allowance?

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