---
title: "STAT 516 Homework 1"
author: "Your classmate"
date: "01/28/2026"
format: html
editor: visual
---

```{r}
hr <- c(80, 118, 92, 84, 78, 84, 
        76, 82, 76, 88, 108, 90, 
        90, 90, 86, 70, 68, 46, 
        98, 56, 84, 80, 78, 66, 60)
```

# 1)

```{r}
qqnorm(hr)
qqline(hr)
```

# 2)

```{r}
xbar <- mean(hr)
sn2 <- var(hr)
sn <- sd(hr)
```

The mean of the sample is $\bar X_n = `r xbar`$, the variance is $S_n^2 = `r round(sn2, 4)`$, and the standard deviation is $S_n = `r round(sn, 4)`$.

# 3)

For a sample with a known standard deviation, the confidence interval of the mean is found by the formula $$
(1-\alpha)100\% \text{ C.I.} = 
\bar X_n \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}
$$

```{r}
alpha <- 1 - 0.99;
n <- length(hr);
sigma <- 16
MoE <- qnorm(1-alpha/2) * sigma / sqrt(n)
```

The $99\% \text{ C.I.}$ of the sample ($\sigma = 16$) is $`r xbar` \pm `r round(MoE, 4)`$.

# 4)

For a sample with an unknown standard deviation (i.e., using $S_n$ to estimate $\sigma$), the confidence interval of the mean is found by the formula $$
(1-\alpha)100\% \text{ C.I.} = 
\bar X_n \pm t_{n-1,\ \alpha/2}\frac{S_n}{\sqrt{n}}
$$

```{r}
alpha <- 1 - 0.99
sn <- sd(hr)
MoE <- qt(1-alpha/2, n-1)*(sn/sqrt(n))
```

The $99\% \text{ C.I.}$ of the sample is $`r xbar` \pm `r round(MoE, 4)`$.

# 5)

The minimum number of samples needed to obtain a desired margin of error, $M^*$, is determined by the formula $$
n = \Bigg\lceil \bigg(z_{\alpha/2}\frac{\sigma}{M^*} \bigg)^2 \Bigg\rceil
$$

```{r}
MoEs <- 2
alpha <- 1 - 0.95
sigma <- sd(hr)
n1 <- ceiling( (qnorm(1-alpha/2) * (sn/MoEs)) ^2 )
```

Using $S_n = `r round(sn, 4)`$ to estimate $\sigma$, we find that number of samples needed is at least $n =`r n1`$.

# 6)

## i)

$$
H_0: \mu \le 80 \text{ bpm}
$$ $$
H_1: \mu > 80 \text{ bpm}
$$

## ii)

$$
T_{stat} = \frac{\bar X_n - \mu_0}{S_n /\sqrt{n}}\\
$$

```{r}
mu0 <- 80
sn <- sd(hr)
ts <- (xbar - mu0) / (sn/sqrt(n) )
```

$$
T_{stat} = `r ts`
$$

## iii)

For a one-sided test, $$
\text{Critical Value} = t_{n-1,\ \alpha}
$$

```{r}
alpha <- 0.05
tc <- qt(1-alpha, n-1)
```

$$
t_{n-1,\ \alpha} = `r tc`
$$

## iv)

We reject $H_0$ if $T_{stat} > t_{n-1,\ \alpha}$.

```{r}
ts > tc
```

$T_{stat} \ngtr t_{n-1,\ \alpha}$. Therefore, we fail to reject $H_0$ and conclude that the mean heart rate of this population of students is not significantly greater than 80 bpm.

## v)

For a right-tailed test, $$
p = P(T > T_{stat})
$$

```{r}
pval <- pt(-ts, n-1)
```

$$ p = `r pval` $$

# 7)

## i)

$$
H_0: \mu = 82 \text{ bpm}
$$ $$
H_1: \mu \ne 82 \text{ bpm}
$$

## ii)

$$
T_{stat} = \frac{\bar X_n - \mu_0}{S_n /\sqrt{n}}\\
$$

```{r}
mu0 <- 82
ts <- (xbar - mu0) / (sn/sqrt(n) )
```

$$
T_{stat} = `r ts`
$$

## iii)

For a two-sided test, $$
\text{Critical Value} = t_{n-1,\ \alpha/2}
$$

```{r}
alpha <- 0.10
tc <- qt(1-alpha/2, n-1)
```

$$
t_{n-1,\ \alpha/2} = `r tc`
$$

## iv)

We reject $H_0$ if $|T_{stat}| > t_{n-1,\ \alpha/2}$.

```{r}
abs(ts) > tc
```

$|T_{stat}| \ngtr t_{n-1,\ \alpha/2}$.Therefore, we fail to reject $H_0$ and conclude that the mean heart rate of this population of students is not significantly different from 82 bpm.

## v)

For a two-tailed test, $$
p = 2*P(T > |T_{stat}|)
$$

```{r}
pval <- 2*pt(-abs(ts), n-1)
```

$$ p = `r pval` $$

# 8)

For a one-tailed test, $$
n = \Bigg\lceil \sigma^2\bigg(\frac{z_{\alpha}+z_{\beta^*}}{\mu^*-\mu_0} \bigg)^2 \Bigg\rceil
$$

```{r}
sigma <- sd(hr)
alpha <- 0.05
beta <- 1 - 0.90
za <- qnorm(1-alpha)
zb <- qnorm(1-beta)
mus <- 77
mu0 <- 75
n_8 <- ceiling(sigma^2 * (za + zb)^2 / (mus - mu0)^2)
```

Given that the true mean of the sample is at least $77 \text{ bpm}$, the number of samples needed to confirm that $\mu > 75 \text{ bpm}$ at significance level $\alpha = 0.05$ and power $\gamma = 0.90$ is $n = `r n_8`$.

# 9)

For a two-tailed test, $$
n = \Bigg\lceil \sigma^2\bigg(\frac{z_{\alpha/2}+z_{\beta^*}}{\mu^*-\mu_0} \bigg)^2 \Bigg\rceil
$$

```{r}
sigma <- sd(hr)
alpha <- 0.01
beta <- 1 - 0.80
za <- qnorm(1-alpha/2)
zb <- qnorm(1-beta)
mus <- 78+2
mu0 <- 78
n_9 <- ceiling(sigma^2 * (za + zb)^2 / (mus - mu0)^2)
```

Given that the true mean of the sample is at least $2 \text{ bpm}$ from $78 \text{ bpm}$, the number of samples needed to confirm that $\mu \ne 78 \text{ bpm}$ at significance level $\alpha = 0.01$ and power $\gamma = 0.80$ is $n = `r n_9`$.