A Normal Distribution entails:

• the mean, median, and mode are (approximately) equal

• the variance of the data is centered about the mean

• occurrences that are near the mean are more frequent than occurrences farther away from the mean

We will be generating random data for this demonstration

data <- rnorm(1000, mean = 50, sd = 10)

This data is:

• 1000 real numbers

• normally distributed around the mean of 50

• with a variance of 100 (standard deviation of 10)

The Central Limit Theorem states that:

• For a population with an unknown mean \(\mu\), taking random samples of the same size n, where n is sufficiently large:

• the sampling distribution of the sample mean (as our statistic of interest) will be approximately normally distributed

• and the mean of the sampling distribution will approximately equal the mean of the population:

• \[\mu_{\overline{x}} = \frac{\mu + ... + \mu}{n} = \mu\]

• with variance:

• \[\sigma^2_{\hat{X}} = \frac{\sigma^2 + ... + \sigma^2}{n^2} = \frac{\sigma^2}{n}\]

Assuming we don’t know the population mean is 50, if we

• take a random sample of 200 observations of our data

• repeat this process 100 times

we can derive a sampling distribution of the mean of the random samples

• This process yeilded a sampling distribution of the mean with 49.81, which is very close to the actual mean of 50

sample_avgs <- replicate(100, mean(sample(data, 200), replace = TRUE))
avg_sample_avg <- mean(sample_avgs)

• For our randomly generated population data, lets assume we didn’t know the true mean was 50. And, lets say that its been claimed that the population mean is 49, something we want to perform a hypothesis test on.

• 1: Parameter of Interest: \(\mu\)

• 2: \(H_0: \mu_0 = 49\)

• 3: \(H_0: \mu > 49\), an upper one-sided hypothesis test

• 4: Test statistic: \(Z_0 = \frac{\overline{X}-\mu_0}{\frac{\sigma}{\sqrt{n}}}\)

• 5: Let us select \(\alpha\) = 0.05, where \(z_{\alpha}\) = 1.645. If our test statistic is greater than 1.645, we reject \(H_0\)

• 6: Let’s take our random sample, n = 300, and determine the test statistic:

samp_mean <- mean(sample(data, 300))
print((samp_mean - 49)/(10/sqrt(300)))

[1] 2.448078

• With a test statistic \(z_0 = 2.45\), we reject the null hypothesis based on our rejection criteria, as \(z_{alpha} = 1.645 < z_0 = 2.45\)

• We make a plot showing the distribution of sample means if 49 was the true mean, and we show the line that our rejection criteria, \(z_{alpha} = 1.645\) lies.

• Then, we show where our test statistic lies, which is within the rejection region (greater than \(z_{alpha}\))

xval = seq(from = (49 - 4 * (10/sqrt(300))), to = (49 + 4 * (10/sqrt(300))),
    length.out = 1000)
yval = dnorm(xval, mean = 49, sd = (10/sqrt(300)))

critical_value = (1.645) * (10/sqrt(300)) + 49

plot <- plot_ly() %>%
    add_lines(x = xval, y = yval, name = "Distribution if mean is 49") %>%
    add_trace(x = critical_value, y = c(0, max(yval)), mode = "lines",
        name = "z_alpha") %>%
    add_trace(x = (samp_mean), y = c(0, max(yval)), mode = "lines",
        name = "test statistic")

plot

• There is a probability we could come to the wrong conclusion from the particular random sample we took, and this probability is determined by our rejection criteria

• Type I Error: We reject the null hypothesis when it is actually true

• This occurs if our test statistic falls within the rejection region

• The value for \(\alpha\) we chose can be interpreted as the probability of having a type I error (for \(\alpha = 0.05\), probability is 5%)

• example: if the true mean was 49, and we happened to get the test statistic we did, we would have falsely rejected \(H_0\)

• Type II Error: We fail to reject the null hypothesis when it is actually false. We would have to know the “true” mean to find it!

• This occurs if our test statistic falls within the “acceptance” region of \(H_0\) (technically, the confidence interval, set by \(\alpha\)

• The probability of having a type II error is denoted as \(\beta\), and for an upper one-sided hypothesis test on the mean with variance known, it is calculated as:

\[ \Phi(z_{\alpha} - \frac{(\mu_{true}-\mu_{0})\sqrt{n}}{\sigma})\] > * Because we know the true mean is 50, we can calculate beta:

z_beta = (1.645 - ((50 - 49) * (sqrt(300)/10)))
beta = pnorm(z_beta, mean = 0, sd = 1, lower.tail = TRUE)
print(beta)

[1] 0.4653156

• We have a 46% probability of committing a type II error!