# one mean

*Students found that the average height of a group of 120 randomly chosen female Sewanee students between the ages of 20 and 29 was \(\bar{x} = 63.8\) inches, with a sample standard deviation of \(s = 2.06\) inches. Does that differ significantly from the heights of all American women in the same age group?*

references:

- Anthropometric Reference Data for Children and Adults: United States, 2011–2014, CDC

`library(tidyverse)`

## observed sample statistic

```
x.bar.observed <- 63.8
s.observed <- 2.06
alpha <- 0.05
```

# simulation

The CDC reports that the average height of American women between the ages of 20 and 29 is 64.1 inches, with a standard deviation of 2.25 inches.

*Design an experiment*:

generate 120 representative heights of American women in this age range … report the average height of the sample

```
n <- 120 # sample size
mu <- 64.1 # mean height (in.)
sigma <- 2.25 # standard deviation (in.)
height120 <- function(){
samp <- rnorm(n, mean = mu, sd = sigma)
x.bar <- mean(samp)
return(x.bar)
}
```

repeat 10 times

`replicate(10, height120())`

```
## [1] 63.78992 64.46217 64.37326 63.98532 64.30370 63.81543 64.09789
## [8] 63.99719 63.87410 63.90026
```

## simulated sampling distribution of \(\bar{x}\)

Assume that the population standard deviation of the heights of these women is known to be \(\sigma\).

**Verify requirements**:

We are assuming that the distribution of heights of American women in this age range is (at least approximately) normal, which means that the sampling distribution of \(\bar{x}\) is also normal, and in any case the sample size \(n = 120\) is greater than 30, so the sampling distribution of \(\bar{x}\) would be approximately normal for that reason as well.

**Expected result**:

Therefore, we would expect the sampling distribution of \(\bar{x}\) to be (1) approximately normal, with (2) mean \[\mu_{\bar{x}} = \mu = 64.1\] and standard error \[{SE}_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = 0.2054\]

```
n.trials <- 1e4
df1 <- data.frame(x.bar = replicate(n.trials, height120()))
str(df1)
```

```
## 'data.frame': 10000 obs. of 1 variable:
## $ x.bar: num 64.4 63.9 64.1 64.1 64.2 ...
```

`sampling.distribution1(df1)`