Lab Homework 1

1.1

Natural Response Variables =

1. Attitude toward gun control
2. Heart Disease
3. Vote for President

Explanatory Variables =

1. Gender, mother’s education
2. Blood pressure, cholesterol level

3. Race, religion, annual income

1.3

1. \(Binomial\); \(n=100\); \(\pi=0.25\)

2. If \(\mu=n\pi=25\) and \(\sigma=\sqrt{n\pi(1-\pi)}=4.33\), then 50 correct responses is surprising, because 50 is \(z=(50-25)/4.33=5.8\) standard deviations above the mean.

1.6

1. Since there are only 3 independent observations (\(y_1+ y_2+ y_3=n\)), \(n_1,n_2,n_3= 3\). If you know \(y_1\) and \(y_2\), you can calculate \(y_3\) The sum of the observed frequencies must equal the total number of observations. This is because the frequencies represent counts of occurrences of each genotype in the sample. where \(n = 3\). Given \(y_1\) and \(y_2\), you can solve for \(y_3\) as: \(y_3 = n - (y_1 + y_2)\)

2. \(300, 210, 201, 120, 111, 102, 030, 021, 012, 003\)

3. \(Binomial~(\pi=.25).\)

1.9

library(exactci)

## Warning: package 'exactci' was built under R version 4.3.3

## Loading required package: ssanv

## Loading required package: testthat

## Warning: package 'testthat' was built under R version 4.3.3

binom.exact(60, 100, conf.level=0.95)

## 
##  Exact two-sided binomial test (central method)
## 
## data:  60 and 100
## number of successes = 60, number of trials = 100, p-value = 0.05689
## alternative hypothesis: true probability of success is not equal to 0.5
## 95 percent confidence interval:
##  0.4972092 0.6967052
## sample estimates:
## probability of success 
##                    0.6

Let \(\pi\) = the population proportion obtaining greater relief with the new analgesic.\(H_0: \pi=0.50;~H_A:\pi\neq 0.50\) For \(H_0\): \(\pi=0.50,~z=2.00,~P-value=0.046.\) We reject the null hypothesis at the 5% significance level. This means that there is evidence to conclude that the probability of greater relief is different between the two analgesics.

x <- 60; n <- 100
alpha <- 0.05; pi.hat <- x/n
var.wald <- pi.hat*(1-pi.hat)/n
lower <- pi.hat - qnorm (p=1-alpha/2)*sqrt(var.wald)
upper <- pi.hat + qnorm (p=1-alpha/2)*sqrt(var.wald)
round(data.frame(lower, upper), 4)

##   lower upper
## 1 0.504 0.696

The 95% confidence interval is approximately \((0.504, 0.696).\) Meaning that we are 95% confident that the true probability of a woman experiencing greater relief with the new analgesic lies between 0.504 and 0.696. This interval suggests that the new analgesic may provide greater relief for a majority of women, but there is still some uncertainty about the exact probability.

1.11

Report on Statistical Inference for the Proportion of American Adults Believing in Heaven

Introduction

The General Social Survey (GSS) conducted a poll in which 1,158 American adults were asked whether they believe in heaven. A total of 86% of respondents answered “yes.” This report aims to use this sample data to estimate the population proportion of American adults who believe in heaven and to conduct a hypothesis test to assess whether the observed sample proportion is consistent with a hypothesized population proportion.

Objective

1. Estimate the true proportion of American adults who believe in heaven.
2. Conduct a hypothesis test to assess whether the true proportion differs from a hypothesized proportion (for example, \(p_0 = 0.80\)).

Methodology

1. Point Estimate

The sample proportion (\(\hat{p}\)) of America is \(\hat{p} = 0.8\)

2. Confidence Interval

To construct a 95% confidence interval for the true population proportion \(p\), we use the formula:

\(\hat{p} \pm Z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\) Where: - \(\hat{p} = 0.86\) (sample proportion) - \(n = 1158\) (sample size) - \(Z_{\alpha/2} \approx 1.96\) for a 95% confidence level Substituting the values:

\(\text{Standard Error} = \sqrt{\frac{0.86 \times (1 - 0.86)}{1158}} = \sqrt{\frac{0.1204}{1158}} \approx 0.0103\) \(\text{Confidence Interval} = 0.86 \pm 1.96 \times 0.0103 \approx 0.86 \pm 0.0202\) \(\text{Confidence Interval} \approx (0.84, 0.88)\)

The 95% confidence interval for the population proportion \(p\) is approximately \((0.84, 0.88)\).

3. Hypothesis Test

Now, we will test the hypothesis that the true population proportion is \(p_0 = 0.80\).

• Null Hypothesis

(\(H_0\)): \(p = 0.80\)

• Alternative Hypothesis

(\(H_A\)): \(p \neq 0.80\) (two-sided test)

The test statistic \(Z\)is calculated as: $ Z = $ Substituting the values: \(Z = \frac{0.86 - 0.80}{\sqrt{\frac{0.80 \times 0.20}{1158}}} = \frac{0.06}{\sqrt{\frac{0.16}{1158}}} = \frac{0.06}{0.0117} \approx 5.13\)——— The corresponding P-value for \(Z = 5.13\) is extremely small (less than 0.0001).

Decision

Since the P-value is much smaller than the standard significance level of 0.05, we reject the null hypothesis \(H_0\). This suggests strong evidence that the true population proportion of American adults who believe in heaven is different from 0.80.

Conclusion

Based on the survey data, the estimated proportion of American adults who believe in heaven is 0.86, with a 95% confidence interval of approximately \((0.840, 0.880)\). Additionally, a hypothesis test strongly rejects the null hypothesis that the true proportion is 0.80, indicating that the proportion is significantly higher.This analysis provides strong statistical evidence that a substantial majority of American adults believe in heaven, with the true proportion likely lying between 84.0% and 88.0%.

1.15

1. For a binomial random variable \(Y \sim \text{Binomial}(n, \pi)\):

• The mean \(\mu_Y\) is \(n\pi\).
• The variance \(\sigma_Y^2\) is \(n\pi(1 - \pi)\).
• The standard deviation \(\sigma_Y\) is \(\sqrt{n\pi(1 - \pi)}\). Since \(\hat{\pi} = \frac{Y}{n}\), the standard deviation of \(\hat{\pi}\) can be derived from the standard deviation of \(Y\). Using the rule that if \(Z = cY\) where \(c\) is a constant, then the standard deviation \(\sigma_Z = c\sigma_Y\). Here, \(\hat{\pi} = \frac{1}{n}Y\), so: \(\sigma(\hat{\pi}) = \frac{1}{n} \times \sigma(Y) = \frac{1}{n} \times \sqrt{n\pi(1 - \pi)} = \sqrt{\frac{\pi(1 - \pi)}{n}}\) Thus, the standard deviation of \(\hat{\pi}\) is: \(\sigma(\hat{\pi}) = \sqrt{\frac{\pi(1 - \pi)}{n}}\)

2. Estimating \(\pi\) is more precise when \(\pi\) is near 0 or 1 because the standard deviation of the sample proportion \(\hat{\pi}\) is smaller, meaning less variability in the estimate. Estimation of \(\pi\) around 0.50 is less precise because the standard deviation is maximized, meaning greater variability in the sample proportion \(\hat{\pi}\).