/*! 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 9 In Exercises 6.9 and 6.10 , indi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 6.9 and 6.10 , indicate whether the Central Limit Theorem applies so that the sample proportions follow a normal distribution. In each case below, is the sample size large enough so that the sample proportions follow a normal distribution? (a) \(n=500\) and \(p=0.1\) (b) \(n=25\) and \(p=0.5\) (c) \(n=30\) and \(p=0.2\) (d) \(n=100\) and \(p=0.92\)

Short Answer

Expert verified
The Central Limit Theorem applies to scenarios (a) and (b), where both yielded more than 10 successes and failures. However, it does not apply to scenarios (c) and (d) as they do not meet the minimum success and failure criteria.

Step by step solution

01

Test Sample Size and Proportion for (a)

Multiply the sample size \(n=500\) and the probability \(p=0.1\). Calculate \(np = 500 * 0.1 = 50\). Multiply the sample size by \(1-p\) to get \(n(1-p) = 500(1-0.1) = 450\). Since both are greater than 10, the Central Limit Theorem applies and the distribution can be considered normal.
02

Test Sample Size and Proportion for (b)

Multiply the sample size \(n=25\) and the probability \(p=0.5\). Calculate \(np = 25 * 0.5 = 12.5\). Similarly, calculate \(n(1-p) = 25(1-0.5) = 12.5\). Since both are greater than 10, the Central Limit Theorem applies and the distribution can be considered normal.
03

Test Sample Size and Proportion for (c)

For \(n=30\) and \(p=0.2\), \(np = 30 * 0.2 = 6\) and \(n(1-p) = 30(1-0.2) = 24\). The first calculation resulted in a number less than 10, which means that the Central Limit Theorem does not apply and we cannot assume a normal distribution.
04

Test Sample Size and Proportion for (d)

For \(n=100\) and \(p=0.92\), calculate \(np = 100 * 0.92 = 92\) and \(n(1-p) = 100(1-0.92) = 8\). Again, since one of the results is less than 10, we cannot apply the Central Limit Theorem, and the distribution may not be normal.

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 91Ó°ÊÓ!

Key Concepts

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

Sample Proportions
Understanding sample proportions is essential for deciphering the success of an action in a larger population. Suppose we're interested in the proportion of voters in a city who support a certain candidate. By collecting data from a simple random sample of the population, we obtain the sample proportion. This value, designated as 'p-hat' (\( \bar{p} \bar{p} \bar{p} \bar{p} \bar{p} \bar{p} \bar{p} \bar{p} \bar{p} \)), acts as an estimate of the true population proportion 'p'.

When we gather data from various samples, the sample proportions themselves have their own distribution, which is centered around the true population proportion. The shape of this distribution depends on the sample size and the true population proportion. If certain conditions are met, the Central Limit Theorem helps us approximate this distribution with a normal distribution which is incredibly useful for making inferences about the population.
Normal Distribution
The normal distribution, often called the bell curve due to its shape, is a continuous probability distribution that is symmetrical around the mean. Many natural phenomena tend to exhibit a normal distribution, given a large enough sample size.

The Central Limit Theorem (CLT) states that the sampling distribution of the sample mean (or proportion) will approach a normal distribution as the sample size increases, regardless of the shape of the population distribution. This is a powerful tool because it allows statisticians to make predictions and inferences about population parameters using the normal distribution, even when the population itself is not normally distributed. However, for sample proportions, the rule of thumb is that both 'np' and 'n(1-p)' must be greater than 10 for the normal approximation to be reliable.
Sample Size
In the realm of statistics, sample size plays a pivotal role in the accuracy of estimations and predictions. The sample size affects the variability of estimates: generally, a larger sample reduces the margin of error and yields more precise estimates. However, increasing the sample size has diminishing returns; doubling the sample size does not halve the margin of error but rather reduces it by a factor of the square root of two.

The Central Limit Theorem also hinges on 'sufficiently large' sample sizes to ensure normal distribution of sample means or proportions. The definition of 'sufficient' can vary, but a common benchmark is that a sample size is large enough if 'np' and 'n(1-p)' are both greater than 10. This rule helps to determine if the CLT can be applied to approximate a normal distribution for sample proportions, further supporting accurate probabilistic modeling and hypothesis testing.
Probability
Probability is the mathematical language we use to quantify uncertainty. It measures the likelihood of an event occurring, and it ranges from 0 (impossible event) to 1 (certain event). Understanding probabilities allows us to make informed decisions based on potential risks and benefits.

In statistics, the probability of an event can be estimated using the relative frequency of outcomes from a well-defined experimental procedure or random sample. With respect to the Central Limit Theorem, probability plays a crucial role in determining how 'unusual' a sample statistic is when compared to a population parameter. For instance, we can calculate the probability that a sample proportion falls within a certain range of the true population proportion. Utilizing the normal distribution, which the CLT provides as an approximation method, we can easily compute these probabilities using standard normal distribution tables or software, allowing us to test hypotheses and make predictions with more confidence.

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

We examine the effect of different inputs on determining the sample size needed. Find the sample size needed to give a margin of error within ±3 with \(99 \%\) confidence. With \(95 \%\) confidence. With \(90 \%\) confidence. Assume that we use \(\tilde{\sigma}=30\) as our estimate of the standard deviation in each case. Comment on the relationship between the sample size and the confidence level desired.

Find a \(95 \%\) confidence interval for the mean two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the t-distribution and the formula for standard error. Compare the results. Mean price of a used Mustang car online, in \$1000s, using data in MustangPrice with \(\bar{x}=15.98\), \(s=11.11,\) and \(n=25\)

In Exercise \(6.107,\) we see that plastic microparticles are contaminating the world's shorelines and that much of the pollution appears to come from fibers from washing polyester clothes. The same study referenced in Exercise 6.107 also took samples from ocean beaches. Five samples were taken from each of 18 different shorelines worldwide, for a total of 90 samples of size \(250 \mathrm{~mL}\). The mean number of plastic microparticles found per \(250 \mathrm{~mL}\) of sediment was 18.3 with a standard deviation of 8.2 . (a) Find and interpret a \(99 \%\) confidence interval for the mean number of polyester microfibers per \(250 \mathrm{~mL}\) of beach sediment. (b) What is the margin of error? (c) If we want a margin of error of only ±1 with \(99 \%\) confidence, what sample size is needed?

Use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. Give the best estimate for \(\mu_{1}-\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=10.1, s_{1}=2.3, n_{1}=50\) and \(\bar{x}_{2}=12.4, s_{2}=5.7, n_{2}=50 .\)

Autism and Maternal Antidepressant Use A recent study \(^{41}\) compared 298 children with Autism Spectrum Disorder to 1507 randomly selected control children without the disorder. Of the children with autism, 20 of the mothers had used antidepressant drugs during the year before pregnancy or the first trimester of pregnancy. Of the control children, 50 of the mothers had used the drugs. (a) Is there a significant association between prenatal exposure to antidepressant medicine and the risk of autism? Test whether the results are significant at the \(5 \%\) level. (b) Can we conclude that prenatal exposure to antidepressant medicine increases the risk of autism in the child? Why or why not? (c) The article describing the study contains the sentence "No increase in risk was found for mothers with a history of mental health treatment in the absence of prenatal exposure to selective serotonin reuptake inhibitors [antidepressants]." Why did the researchers conduct this extra analysis?

See all solutions

Recommended explanations on Math Textbooks

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.