Question

Assume that I asked heights of 40 female students. It turns out that average height and standard deviation of ladies in our class is 164 cm and 9 cm, respectively. For all computations, assume that the heights follow a normal distribution. (STATISTICS)

a) Write a function in R that calculates the confidence interval for given ?, X̄, s and n.

b) Write a function in R that calculates the prediction interval for given ?, X̄, s and n.

c) Write a function in R that conducts a hypothesis test. The function’s output will be “reject” or “don’t reject” using the input values of ?, X̄, s and n.

Answer 1

Here, n = 40

= 164

s = 9

a) To calculate confidence interval, we can use the below function

Below solution if for 5% significance. Can be charged as per a value

> a <- 0.05 
> x <- 164
> s <- 9
> n <- 40
> error <- qnorm(1-a/2)*s/sqrt(n)
> left <- a-error
> right <- a+error
> left
[1] 161.2109
> right
[1] 166.7891

b) Prediction interval is given by The function in R will be written as:-

> a <- 0.05
> x <- 164

> s <- 9

> error <- qnorm(1-a/2)*s

> left <- a-error

> right <- a+error

> left [1]

146.36

> right [1]

181.64

c) Function for two-tailed hypothesis test

The null hypothesis is that μ = 170. We begin with computing the test statistic.

> x = 164 # sample mean
> mu = 170 # hypothesized value
> s = 9 # sample standard deviation
> n = 40 # sample size
> z = (xbar−mu)/(s/sqrt(n))
> z                      # test statistic
[1] -4.21637

We then compute the critical values at .05 significance level.

> a = .05
> z.half.alpha = qnorm(1−a/2)
> c(−z.half.alpha, z.half.alpha)
[1] −1.9600  1.9600

> If −z.half.alpha <= z <=  z.half.alpha, "Do No Reject", "Reject"

[1] Reject