/*! 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 45 A random sample of 150 recent do... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 45

A random sample of 150 recent donations at a certain 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 heen nsed?

Short Answer

Expert verified
The proportion of type A donations significantly differs from 40% at both 0.01 and 0.05 significance levels.

Step by step solution

01

Define the Hypotheses

To determine if the actual percentage of type A blood donations differs from 40%, define the null and alternative hypotheses. The null hypothesis (\(H_0\)) is that the proportion of type A donations is \(p_0 = 0.40\). The alternative hypothesis (\(H_1\)) is that the proportion of type A donations is not 0.40, \(p eq 0.40\).
02

Calculate Sample Proportion

Calculate the sample proportion \(\hat{p}\) of type A blood donations. The formula for sample proportion is \(\hat{p} = \frac{x}{n}\), where \(x\) is the number of type A donations and \(n\) is the total number of donations. Thus, \(\hat{p} = \frac{82}{150} = 0.547\).
03

Test Statistic Calculation

Calculate the test statistic using the formula for a proportion test: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] Substituting the values, we get \[ z = \frac{0.547 - 0.40}{\sqrt{\frac{0.40(0.60)}{150}}} \approx 3.71 \]
04

Determine Critical Value and Decision Rule

For a two-tailed test at a significance level of 0.01, find the critical z-values. The critical z-values are \(-2.576\) and \(2.576\). If the test statistic falls outside this range, reject the null hypothesis.
05

Compare Test Statistic to Critical Value

The calculated z-value is approximately 3.71, which is greater than 2.576. Therefore, the test statistic falls in the rejection region, and we reject the null hypothesis.
06

Conclusion at 0.01 Significance Level

Since the null hypothesis is rejected at the 0.01 significance level, there is evidence to suggest that the proportion of type A donations differs from 40%.
07

Consider Different Significance Level

If the significance level was 0.05, the critical z-values would be \(-1.96\) and \(1.96\). The z-value of 3.71 still falls outside this range, so the null hypothesis would also be rejected at this level.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Proportion Test
A proportion test helps us determine whether the proportion of a certain characteristic in a sample differs from a specified value in the population. In our case, we want to know if the proportion of type A blood donations is different from the hypothesized population proportion of 40%. This test is crucial when dealing with categorical data, like blood types, where the outcome can be binary (type A or not type A).
  • First, we establish our null hypothesis (\(H_0\)): the population proportion is 0.40.
  • Next, the alternative hypothesis (\(H_1\)): the population proportion is not 0.40.
  • The sample proportion is simply the number of observations in our sample that fit our category of interest (in this case, type A blood), divided by the total number of observations.
This method allows us to use sample data to infer things about the general population.
Significance Level
The significance level is a threshold we set to decide if an observed effect is statistically significant. It is denoted by \(\alpha\) and represents the probability of rejecting the null hypothesis when it is actually true. A lower significance level means a stricter criterion to declare that a result is statistically significant.
  • A common significance level is 0.05, meaning there's a 5% chance of a Type I error (incorrectly rejecting a true null hypothesis).
  • In this exercise, we're using both \(\alpha = 0.01\) and \(\alpha = 0.05\) to see how our results compare under strict and more lenient criteria.
If the p-value calculated from our test statistic is less than the chosen significance level, it suggests the sample provides enough evidence against the null hypothesis.
Critical Value
Critical values are a key part of hypothesis testing. They define the cutoff points in a probability distribution, helping to decide whether the null hypothesis can be rejected or not. In our test of proportions, we use a z-distribution since we're assuming a normal distribution for large samples.
  • For a two-tailed test at \(\alpha = 0.01\), the critical z-values are \(-2.576\) and \(2.576\).
  • At \(\alpha = 0.05\), these values are \(-1.96\) and \(1.96\).
If our calculated z-statistic is beyond these critical values, we reject the null hypothesis. This means there is a statistically significant difference between the observed and hypothesized proportions.
Z-Test
A z-test is a statistical test used to determine if there is a significant difference between sample and population proportions. Here's how it works in our context:
  • Calculate the sample proportion (\(\hat{p}\)).
  • Use the formula for the z-test statistic: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]
  • Compare this calculated z-value to the critical z-value to determine the outcome.
In our example, the z-value of 3.71 significantly exceeded the critical values for both \(alpha = 0.01\) and \(alpha = 0.05\). This meant we rejected the null hypothesis, suggesting that the true percentage of type A donors is not 40%. The z-test is a powerful tool for making strong inferences about population parameters.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Each of a group of 20 intermediate tennis players is given two rackets, one having nylon strings and the other synthetic gut strings. After several weeks of playing with the two rackets, each player will be asked to state a preference for one of the two types of strings. Let \(p\) denote the proportion of all such players who would prefer gut to nylon, and let \(X\) be the number of players in the sample who prefer gut. Because gut strings are more expensive, consider the null hypothesis that at most \(50 \%\) of all such players prefer gut. We simplify this to \(H_{0}: p=.5\), planning to reject \(H_{0}\) only if sample evidence strongly favors gut strings.a. Is a significance level of exactly 05 achievable? If not, what is the largest \(\alpha\) smaller than \(.05\) that is achievable? b. If \(60 \%\) of all enthusiasts prefer gut, calculate the probability of a type II error using the significance level from part (a). Repeat if \(80 \%\) of all enthusiasts prefer gut. c. If 13 out of the 20 players prefer gut, should \(H_{0}\) be rejected using the significance level of (a)?

The article "Effects of Bottle Closure Type on Consumer Perception of Wine Quality" (Amer. \(J\). of Enology and Viticulfure, 2007: 182-191) reported that in a sample of 106 wine consumers, \(22(20.8 \%)\) thought that screw tops were an acceptable substitute for natural corks. Suppose a particular winery decided to use screw tops for one of its wines unless there was strong evidence to suggest that fewer than \(25 \%\) of wine consumers found this acceptable. a. Using a significance level of .10, what would you recommend to the winery? b. For the hypotheses tested in (a), describe in context what the type I and II errors would be, and say which type of error might have been committed.

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 \(40 \mathrm{mph}\) 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 values of \(\bar{x}\) are more contradictory to \(H_{0}\) than \(117.2\), what is the \(P\)-value in this case, and what conclusion is appropriate if \(\alpha=.10\) ? c. What is the probability that the new design is not implemented when its true average braking distance is actually \(115 \mathrm{ft}\) and the test from part (b) is used?

The calibration of a scale is to be checked by weighing a \(10-\mathrm{kg}\) test specimen 25 times. Suppose that the results of different weighings are independent of one another and that the weight on each trial is normally distributed with \(\sigma=.200 \mathrm{~kg}\). Let \(\mu\) denote the true average weight reading on the scale. a. What hypotheses should be tested? b. With the sample mean itself as the test statistic, what is the \(P\)-value when \(\bar{x}=9.85\), and what would you conclude at significance level .01? c. For a test with \(\alpha=.01\), what is the probability that recalibration is judged unnecessary when in fact \(\mu=\) 10.1? When \(\mu=9.8\) ?

Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorable, so a potential customer has decided to go ahead with a purchase arrangement unless it can be conclusively demonstrated that the true average lifetime is smaller than what is advertised. A random sample of 50 bulbs was selected, the lifetime of each bulb determined, and the appropriate hypotheses were tested using Minitab, resulting in the accompanying output. \(\begin{array}{lrrrrrr}\text { Variable } & N & \text { Mean } & \text { StDev } & \text { SE Mean } & \text { Z } & \text { P-Value } \\ \text { lifetime } 50 & 738.44 & 38.20 & 5.40 & -2.14 & 0.016\end{array}\) What conclusion would be appropriate for a significance level of \(.05\) ? A significance level of .01? What significance level and conclusion would you recommend?
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.