STAT 516 Homework 1

Author

Your classmate

Published

January 28, 2026

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)

qqnorm(hr)
qqline(hr)

2)

xbar <- mean(hr)
sn2 <- var(hr)
sn <- sd(hr)

The mean of the sample is ${\bar{X}}_{n} = 81.12$ , the variance is $S_{n}^{2} = 238.6933$ , and the standard deviation is $S_{n} = 15.4497$ .

3)

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

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

The $99 % C.I.$ of the sample ( $σ = 16$ ) is $81.12 \pm 8.2427$ .

4)

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

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

The $99 % C.I.$ of the sample is $81.12 \pm 8.6424$ .

5)

The minimum number of samples needed to obtain a desired margin of error, $M^{*}$ , is determined by the formula $n = ⌈ (z_{α / 2} \frac{σ}{M^{*}})^{2} ⌉$

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

Using $S_{n} = 15.4497$ to estimate $σ$ , we find that number of samples needed is at least $n = 230$ .

6)

i)

$H_{0} : μ \leq 80 bpm$ $H_{1} : μ > 80 bpm$

ii)

$T_{s t a t} = \frac{{\bar{X}}_{n} - μ_{0}}{S_{n} / \sqrt{n}}$

mu0 <- 80
sn <- sd(hr)
ts <- (xbar - mu0) / (sn/sqrt(n) )

$T_{s t a t} = 0.3624665$

iii)

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

alpha <- 0.05
tc <- qt(1-alpha, n-1)

$t_{n - 1, α} = 1.7108821$

iv)

We reject $H_{0}$ if $T_{s t a t} > t_{n - 1, α}$ .

ts > tc

[1] FALSE

$T_{s t a t} ≯ t_{n - 1, α}$ . 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, $p = P (T > T_{s t a t})$

pval <- pt(-ts, n-1)

$p = 0.3600879$

7)

i)

$H_{0} : μ = 82 bpm$ $H_{1} : μ \neq 82 bpm$

ii)

$T_{s t a t} = \frac{{\bar{X}}_{n} - μ_{0}}{S_{n} / \sqrt{n}}$

mu0 <- 82
ts <- (xbar - mu0) / (sn/sqrt(n) )

$T_{s t a t} = - 0.2847951$

iii)

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

alpha <- 0.10
tc <- qt(1-alpha/2, n-1)

$t_{n - 1, α / 2} = 1.7108821$

iv)

We reject $H_{0}$ if $| T_{s t a t} | > t_{n - 1, α / 2}$ .

abs(ts) > tc

[1] FALSE

$| T_{s t a t} | ≯ t_{n - 1, α / 2}$ .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, $p = 2 * P (T > | T_{s t a t} |)$

pval <- 2*pt(-abs(ts), n-1)

$p = 0.7782442$

8)

For a one-tailed test, $n = ⌈ σ^{2} (\frac{z_{α} + z_{β^{*}}}{μ^{*} - μ_{0}})^{2} ⌉$

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 $77 bpm$ , the number of samples needed to confirm that $μ > 75 bpm$ at significance level $α = 0.05$ and power $γ = 0.90$ is $n = 512$ .

9)

For a two-tailed test, $n = ⌈ σ^{2} (\frac{z_{α / 2} + z_{β^{*}}}{μ^{*} - μ_{0}})^{2} ⌉$

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 $2 bpm$ from $78 bpm$ , the number of samples needed to confirm that $μ \neq 78 bpm$ at significance level $α = 0.01$ and power $γ = 0.80$ is $n = 697$ .