/*! 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 3 Suppose that \(90 \%\) of orange... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that \(90 \%\) of orange tabby cats are male. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of sample proportions of random samples of size 30 is left skewed. (b) Using a sample size that is 4 times as large will reduce the standard error of the sample proportion by one-half. (c) The distribution of sample proportions of random samples of size 140 is approximately normal. (d) The distribution of sample proportions of random samples of size 280 is approximately normal.

Short Answer

Expert verified
(a) False, (b) True, (c) True, (d) True.

Step by step solution

01

Understand Statement (a)

Statement (a) claims that the distribution of sample proportions of random samples of size 30 is left skewed. Generally, the Central Limit Theorem states that the sampling distribution of the sample proportion becomes approximately normal if both \( n \cdot p \geq 10 \) and \( n \cdot (1-p) \geq 10 \) are satisfied, where \( n \) is the sample size and \( p \) is the population proportion. Here, \( n = 30 \) and \( p = 0.9 \). So, \( n \cdot p = 30 \cdot 0.9 = 27 \) and \( n \cdot (1-p) = 30 \cdot 0.1 = 3 \). Since \( n \cdot (1-p) < 10 \), the distribution is not approximately normal, and because \( p \) is close to 1 (a small minority of observations for the other category), the distribution is right-skewed, not left-skewed. Therefore, the statement is False.
02

Evaluate Statement (b)

Statement (b) claims that using a sample size that is 4 times as large will reduce the standard error of the sample proportion by one-half. The standard error for the sample proportion \( \hat{p} \) is given by \( \sqrt{\frac{p(1-p)}{n}} \). If the sample size \( n \) is increased by a factor of 4, i.e., \( n' = 4n \), then the new standard error is \( \sqrt{\frac{p(1-p)}{4n}} = \frac{1}{2} \cdot \sqrt{\frac{p(1-p)}{n}} \). Therefore, increasing the sample size by a factor of 4 indeed reduces the standard error by one-half. Thus, the statement is True.
03

Examine Statement (c)

Statement (c) indicates that the distribution of sample proportions of random samples of size 140 is approximately normal. Using \( p = 0.9 \), we calculate \( n \cdot p = 140 \cdot 0.9 = 126 \) and \( n \cdot (1-p) = 140 \cdot 0.1 = 14 \). Both values exceed 10, fulfilling the condition for the approximation to normality as per the Central Limit Theorem. Hence, the distribution is approximately normal. Therefore, the statement is True.
04

Analyze Statement (d)

Statement (d) suggests that the distribution of sample proportions of random samples of size 280 is approximately normal. Calculate \( n \cdot p = 280 \cdot 0.9 = 252 \) and \( n \cdot (1-p) = 280 \cdot 0.1 = 28 \). Both values are much greater than 10, thus satisfying the conditions for the approximation to normality under the Central Limit Theorem. Therefore, the distribution is indeed approximately normal, making the statement True.

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
In statistics, when we talk about sample proportions, we are referring to the fraction or percentage of the sample that has a particular trait or characteristic. For example, if we have a group of 100 cats and 90 of them are orange tabby males, the sample proportion is 0.9 or 90%.

Sample proportions are an important part of inferential statistics because they allow us to make predictions about the entire population. By studying a representative sample, statisticians can estimate characteristics of the whole population without having to survey everyone. This saves time, effort, and resources.

However, it's crucial to ensure that the sample is randomly selected. Random sampling helps avoid bias and increases the likelihood that the sample proportion accurately reflects the population proportion. In the context of the original exercise, we explored whether different samples of a specific size follow the expected distribution.
Sampling Distribution
A sampling distribution is a probability distribution of a statistic obtained from a large number of samples drawn from a specific population. It essentially maps all the possible values that a statistic can assume from different samples of the same size. In the context of our exercise, the sampling distribution refers to the distribution of sample proportions for different sample sizes.

The Central Limit Theorem plays a crucial role in understanding sampling distributions. It tells us that if we have a large enough sample size, the distribution of sample proportions will be approximately normal (bell-shaped), regardless of the shape of the population distribution. This means, for large enough sample sizes, we can use the properties of normal distributions to make inferences about the sample proportions.

In practice, such normal distribution occurs when both the following conditions hold:
  • \( n \cdot p \geq 10 \)
  • \( n \cdot (1 - p) \geq 10 \).
When these conditions are met, the sampling distribution of sample proportions will tend towards a normal distribution as shown in statements (c) and (d) of the original problem.
Normal Distribution
The normal distribution, often called the bell curve, is a continuous probability distribution that is symmetrical around its mean, showing that data near the mean are more frequent in occurrence than data far from the mean. Many real-world variables are distributed normally, or close to it, which is why the normal distribution is such a fundamental concept in statistics.

In terms of our exercise, normal distribution becomes relevant when determining the shape of the sampling distribution of sample proportions. As sample sizes increase, the Central Limit Theorem indicates that the sampling distribution will tend to become normally distributed, provided our earlier conditions ( \( n \cdot p \geq 10 \) and \( n \cdot (1 - p) \geq 10 \)) are satisfied.

This principle helped us evaluate statements concerning sample sizes of 140 and 280, where we found the distribution to be approximately normal due to satisfying these necessary conditions.
Standard Error
The standard error is a measure of the variability or dispersion of a sampling distribution. It indicates how much the sample proportion is expected to vary from the true population proportion due to random sampling. A smaller standard error suggests that the sample proportion is likely to be closer to the population proportion.

Mathematically, the standard error of the sample proportion \( \hat{p} \) is calculated as: \[ SE = \sqrt{\frac{p(1-p)}{n}} \]This formula shows us that the standard error depends on both the population proportion \( p \) and the sample size \( n \).

In the exercise, statement (b) explored how increasing the sample size affects the standard error. By quadrupling the sample size, the standard error was reduced by one-half, illustrating the inverse relationship between sample size and standard error. This means larger samples offer more precise estimates of the population proportion, as we demonstrated with our simple math.

Understanding standard error is essential for assessing the reliability of our statistical estimates.

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

The OpenIntro website occasionally experiments with design and link placement. We conducted one experiment testing three different placements of a download link for this textbook on the book's main page to see which location, if any, led to the most downloads. The number of site visitors included in the experiment was 701 and is captured in one of the response combinations in the following table: $$ \begin{array}{lcc} \hline & \text { Download } & \text { No Download } \\ \hline \text { Position 1 } & 13.8 \% & 18.3 \% \\ \text { Position 2 } & 14.6 \% & 18.5 \% \\ \text { Position 3 } & 12.1 \% & 22.7 \% \\ \hline \end{array} $$ (a) Calculate the actual number of site visitors in each of the six response categories. (b) Each individual in the experiment had an equal chance of being in any of the three experiment groups. However, we see that there are slightly different totals for the groups. Is there any evidence that the groups were actually imbalanced? Make sure to clearly state hypotheses, check conditions, calculate the appropriate test statistic and the p-value, and make your conclusion in context of the data. (c) Complete an appropriate hypothesis test to check whether there is evidence that there is a higher rate of site visitors clicking on the textbook link in any of the three groups.

The United States federal government shutdown of \(2018-2019\) occurred from December 22,2018 until January \(25,2019,\) a span of 35 days. A Survey USA poll of 614 randomly sampled Americans during this time period reported that \(48 \%\) of those who make less than \(\$ 40,000\) per year and \(55 \%\) of those who make \(\$ 40,000\) or more per year said the government shutdown has not at all affected them personally. A \(95 \%\) confidence interval for \(\left(p_{<40 \mathrm{~K}}-p_{\geq 40 \mathrm{~K}}\right),\) where \(p\) is the proportion of those who said the government shutdown has not at all affected them personally, is \((-0.16,0.02) .\) Based on this information, determine if the following statements are true or false, and explain your reasoning if you identify the statement as false. \(^{24}\) (a) At the \(5 \%\) significance level, the data provide convincing evidence of a real difference in the proportion who are not affected personally between Americans who make less than \(\$ 40,000\) annually and Americans who make \(\$ 40,000\) annually. (b) We are \(95 \%\) confident that \(16 \%\) more to \(2 \%\) fewer Americans who make less than \(\$ 40,000\) per year are not at all personally affected by the government shutdown compared to those who make \(\$ 40,000\) or more per year. (c) A \(90 \%\) confidence interval for \(\left(p_{<40 \mathrm{~K}}-p_{\geq 40 \mathrm{~K}}\right)\) would be wider than the (-0.16,0.02) interval. (d) A \(95 \%\) confidence interval for \(\left(p_{\geq 40 \mathrm{~K}}-p_{<40 \mathrm{~K}}\right)\) is (-0.02,0.16) .

Determine if the statements below are true or false. For each false statement, suggest an alternative wording to make it a true statement. (a) As the degrees of freedom increases, the mean of the chi-square distribution increases. (b) If you found \(\chi^{2}=10\) with \(d f=5\) you would fail to reject \(H_{0}\) at the \(5 \%\) significance level. (c) When finding the p-value of a chi-square test, we always shade the tail areas in both tails. (d) As the degrees of freedom increases, the variability of the chi-square distribution decreases.

A local news outlet reported that \(56 \%\) of 600 randomly sampled Kansas residents planned to set off fireworks on July \(4^{t h} .\) Determine the margin of error for the \(56 \%\) point estimate using a \(95 \%\) confidence level. \(^{10}\)

A study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A \(95 \%\) confidence interval for the difference between the proportions of males and females whose favorite color is black \(\left(p_{\text {male }}-p_{\text {female }}\right)\) was calculated to be (0.02,0.06) . Based on this information, determine if the following statements are true or false, and explain your reasoning for each statement you identify as false. \(^{23}\) (a) We are \(95 \%\) confident that the true proportion of males whose favorite color is black is \(2 \%\) lower to \(6 \%\) higher than the true proportion of females whose favorite color is black. (b) We are \(95 \%\) confident that the true proportion of males whose favorite color is black is \(2 \%\) to \(6 \%\) higher than the true proportion of females whose favorite color is black. (c) \(95 \%\) of random samples will produce \(95 \%\) confidence intervals that include the true difference between the population proportions of males and females whose favorite color is black. (d) We can conclude that there is a significant difference between the proportions of males and females whose favorite color is black and that the difference between the two sample proportions is too large to plausibly be due to chance. (e) The \(95 \%\) confidence interval for \(\left(p_{\text {female }}-p_{\text {male }}\right)\) cannot be calculated with only the information given in this exercise.

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.