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Vegetarian college students. Suppose that \(8 \%\) of college students are vegetarians. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of the sample proportions of vegetarians in random samples of size 60 is approximately normal since \(n \geq 30\). (b) The distribution of the sample proportions of vegetarian college students in random samples of size 50 is right skewed. (c) A random sample of 125 college students where \(12 \%\) are vegetarians would be considered unusual. (d) A random sample of 250 college students where \(12 \%\) are vegetarians would be considered unusual. (e) The standard error would be reduced by one-half if we increased the sample size from 125 to 250 .

Short Answer

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

Step by step solution

01

Understanding Normal Approximation Conditions

To determine if the distribution of sample proportions is approximately normal, we use the criteria for normal approximation: a sample proportion is approximately normal if both \( np \geq 10 \) and \( n(1-p) \geq 10 \), where \( n \) is the sample size and \( p \) is the proportion of vegetarians. Here, \( p = 0.08 \) and \( n = 60 \). Calculate \( np = 60 \times 0.08 = 4.8 \) and \( n(1-p) = 60 \times 0.92 = 55.2 \). Therefore, (a) is false because \( np < 10 \).
02

Determine Skewness of Distribution

The same conditions apply here to determine normality. For \( n = 50 \), \( np = 50 \times 0.08 = 4 \) and \( n(1-p) = 50 \times 0.92 = 46 \). Since \( np < 10 \), the distribution is not approximately normal and could be right skewed. Thus, (b) is true.
03

Assess Unusualness with 125 Students

For \( n = 125 \), calculate the expected number and standard error. The expected count is \( np = 125 \times 0.08 = 10 \), just meeting conditions for normality. The standard error is \( \sqrt{\frac{0.08 \times 0.92}{125}} \approx 0.024 \). A sample proportion of \( 12\% \) is \( 0.12 \). Compute the z-score: \( z = \frac{0.12 - 0.08}{0.024} \approx 1.67 \). This is within 2 standard deviations, so (c) is false; it is not unusual.
04

Assess Unusualness with 250 Students

For \( n = 250 \), expected count \( np = 20 \) satisfies normality. Calculate new standard error \( \sqrt{\frac{0.08 \times 0.92}{250}} \approx 0.018 \). For \( 12\% \), compute z-score: \( z = \frac{0.12 - 0.08}{0.018} \approx 2.22 \). As this is slightly above 2 standard deviations, (d) is true; it is considered unusual.
05

Compare Standard Errors with Increased Sample Size

Original standard error for \( n = 125 \) is \( 0.024 \) and for \( n = 250 \) is \( 0.018 \). The standard error decreases with the square root of the sample size, but it is not exactly halved (\( \frac{1}{\sqrt{2}} \approx 0.707 \)). Therefore, (e) is false.

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Key Concepts

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

Normal Approximation
When dealing with statistical inference, it is often important to determine whether a sample proportion distribution can be approximated by a normal distribution. This is known as the normal approximation. To decide if this is possible, we apply the criteria:
  • The condition \( np \geq 10 \)
  • The condition \( n(1-p) \geq 10 \)
These criteria ensure the sample size is large enough so that the sampling distribution of the proportion is approximately normal. In the exercise provided, these conditions were used to assess whether the sample sizes of 60 and 50 might lead to a normal approximation.
However, the calculation \( np = 60 \times 0.08 = 4.8 \) reveals that the condition \( np \geq 10 \) is not met, indicating that the distribution cannot be approximated as normal. Understanding these conditions help in making inferences about population proportions using sample data.
Sample Proportion
The concept of sample proportion is fundamental in statistics when examining samples to make conclusions about a population. The sample proportion represents the fraction of the sample that has a certain characteristic. In our context, if 12% of a sample of students are vegetarians, the sample proportion \( \hat{p} \) is 0.12.
The sample proportion is used in various calculations, including determining the standard error and the z-score.
  • It gives an estimate of the population proportion \( p \).
  • It is crucial in hypothesis testing to check if observed proportions differ from assumed population proportions.
By consistently applying the concept of sample proportion, statisticians can estimate and infer the characteristics of an entire population based on sampled data.
Standard Error
Standard error is a measure of the dispersion or spread of the sample proportion, showing how much variation there is from the actual population proportion. It's calculated using the formula: \[SE = \sqrt{\frac{p(1-p)}{n}}\]where \( p \) is the population proportion, and \( n \) is the sample size.
The standard error decreases as the sample size increases, which represents a reduction in sampling variability.
  • A smaller standard error indicates a more precise estimate of the population proportion.
  • For larger sample sizes, the standard error is smaller, as observed in the calculation where the standard error for 125 students was reduced when the sample was increased to 250 students.
Understanding the standard error is key to assessing how sample statistics approximate the population parameters.
Z-Score
The z-score is a statistical measure that describes a value's position relative to the mean of a group of values. It helps determine how unusual or typical a sample statistic is under a given hypothesis. When comparing sample proportions to the population proportion, the z-score helps in deducing unusualness. It is calculated as:\[z = \frac{\hat{p} - p}{SE}\]where \( \hat{p} \) is the sample proportion, \( p \) is the population proportion, and \( SE \) is the standard error.
A z-score tells us how many standard deviations the sample proportion is from the population proportion.
  • A z-score within the range of roughly \(-2\) to \(2\) suggests that the statistic is not unusual.
  • A higher absolute value of the z-score indicates that the sample proportion is further from what would be expected if the null hypothesis were true.
In the exercise, z-scores were used to interpret if the observed proportion of vegetarians in larger samples was unusual, aiding decisions about statistical significance.

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Most popular questions from this chapter

Is college worth it? Part. I. Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, \(48 \%\) said they decided not to go to college because they could not afford school. 40 (a) A newspaper article states that only a minority of the Americans who decide not to go to college do so because they cannot afford it and uses the point estimate from this survey as evidence. Conduct a hypothesis test to determine if these data provide strong evidence supporting this statement. (b) Would you expect a confidence interval for the proportion of American adults who decide not to go to college because they cannot afford it to include 0.5? Explain.

HIV in sub-Saharan Africa. In July 2008 the US National Institutes of Health announced that it was stopping a clinical study early because of unexpected results. The study population consisted of HIV-infected women in sub-Saharan Africa who had been given single dose Nevaripine (a treatment for HIV) while giving birth, to prevent transmission of HIV to the infant. The study was a randomized comparison of continued treatment of a woman (after successful childbirth) with Nevaripine vs. Lopinavir, a second drug used to treat HIV. 240 women participated in the study; 120 were randomized to each of the two treatments. Twenty-four weeks after starting the study treatment, each woman was tested to determine if the HIV infection was becoming worse (an outcome called virologic failure). Twenty-six of the 120 women treated with Nevaripine experienced virologic failure, while 10 of the 120 women treated with the other drug experienced virologic failure. (a) Create a two-way table presenting the results of this study. (b) State appropriate hypotheses to test for independence of treatment and virologic failure. (c) Complete the hypothesis test and state an appropriate conclusion. (Reminder: verify any necessary conditions for the test.)

True or false, Part. II. 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 \(X^{2}=10\) 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.

Gender and color preference. A 2001 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 ( \(p_{\text {male }}-p\) female \()\) 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. 2 (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.

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