---
title: "Homework 9"
author: "Alizeh Azhar"
date: "10/27/2021"
output:
  pdf_document: default
  html_document: default
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```

## Problem 1: Always, Sometimes, or Never (8 points)

The following statements are always, sometimes, or never true. Mark each statement with Always, Sometimes, or Never, depending on the truth of each statement

1.  The Bernoulli distribution is based on the geometric distribution\
    **Answer**\
    \textcolor{blue}{Always}

2.  For any two events *A* and *B*, *P(A or B)* = *P(A)* $+$ *P(B)*.\
    **Answer**\
    \textcolor{blue}{Sometimes}

3.  *X* and *Y* are two variables. Small values of *X* result in large values of *Y* , and large values of *X* result in small values of *Y* . *X* and *Y* are negatively related.\
    **Answer**\
    \textcolor{blue}{Always}

4.  The standard normal distribution is a discrete distribution.\
    **Answer**\
    \textcolor{blue}{Never}

5.  Box plots are good for investigating the distribution of a categorical variable.\
    **Answer**\
    \textcolor{blue}{Sometimes}

6.  If *X* is right-skewed, the median is less than the mean.\
    **Answer**\
    \textcolor{blue}{Always}

7.  A sample is a subset of a population.\
    **Answer**\
    \textcolor{blue}{Always}

8.  In a valid probability distribution, the sum of the probabilities of the disjoint outcomes must sum to one.\
    **Answer**\
    \textcolor{blue}{Always}

## Problem 2: OIS Exercise 5.19 page 202 (4 points)

Teens were surveyed about cyberbullying, and $54\%$ to $64\%$ reported experiencing cyberbullying ($95\%$ confidence interval). Answer the following questions based on this interval.

(a) A newspaper claims that a majority of teens have experienced cyberbullying. Is this claim supported by the confidence interval? Explain your reasoning.\
    **Answer**\
    \textcolor{blue}{This claim is reasonable, since the entire in- terval lies above 50\%.}

(b) A researcher conjectured that $70\%$ of teens have experienced cyberbullying. Is this claim supported by the confidence interval? Explain your reasoning.\
    **Answer**\
\textcolor{blue}{The value of 70\% lies out- side of the interval, so we have convincing evidence that the researcher's conjecture is wrong.}

(c) Without actually calculating the interval, determine if the claim of the researcher from part (b) would be supported based on a $90\%$ interval. Explain your reasoning.\
    **Answer**\
\textcolor{blue}{A 90\% confidence interval will be narrower than a 95\% confidence interval. Even without calculating the interval, we can tell that 70\% would not fall in the interval, and we would reject the researcher's conjecture based on a 90\% confidence level as well.}

(d) Would a hypothesis test of $H_0 : p = 0.70$ be rejected or fail to rejected, based on the confidence interval? Explain your reasoning.\
    **Answer**\
    \textcolor{blue}{The hypothesis would be rejected. This is because 0.70 is not contained in the 95\% confidence interval.}

## Problem 3: OIS Exercise 6.20, page 226 (4 points)

The United States federal government shutdown of $2018-2019$ occurred from December 22, 2018 to 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 that the government shutdown has not at all affected them personally. A $95\%$ confidence interval for $(p<40k-p\geq40k)$, 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.

(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 (1 pt).\
    **Answer**\
    \textcolor{blue}{False. The interval contains point 0. This indicates that there is no significant difference between this proportion.}

(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 (1 pt).\
    **Answer**\
    \textcolor{blue}{False. We are 95\% confident that $2\%$ more to $16\%$ 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 $(p<40k-p\geq40k)$ would be wider than the $(-0.16, 0.02)$ interval (1 pt).\
    **Answer**\
    \textcolor{blue}{False. The 90\% confidence interval is always narrower than 95\% confidence as the critical value is small for 90\% confidence interval.}

(d) A $95\%$ confidence interval for $(p \geq40k-p<40k)$ is $(-0.02, 0.16)$ (1 pt).\
    **Answer**\
    \textcolor{blue}{True.}

## Problem 4: OIS Exercise 5.32, page 205 (4 pts)

It is believed that nearsightedness affects about $8\%$ of all children. In a random sample of 194 children, 21 are nearsighted. Conduct a hypothesis test for the following question: do these data provide evidence that the $8\%$ value is accurate?\
**Answer**\
\textcolor{blue}{The Hypothesis test for the question is: $$H_0: p=0.08$$ vs $$H_A : p \neq 0.08$$}