Programming Assignment 1

Frequentist Statistics

Group Members: Rene Padilla and Shanmukh Sain Karri

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Prompt:

According to the Bureau of Crime Statistics and Research of Australia, the mean length of imprisonment for motor-vehicle-theft offenders in Australia is 16.7 months. A group of researchers would like to perform a hypothesis test to decide whether the mean length of imprisonment for motor-vehicle-theft offenders in Sydney differs from the national mean in Australia. They have found out that Sydney population standard deviation is 6.0. They have also decided to choose a random sample of size 100 and perform the test at the significance level of 0.05. Suppose that, in reality, the mean length of imprisonment in Sydney is 15.5 months.

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# Importing Libraries
library(stats)

# Defining Given Values
n = 100
alpha = 0.05
mu = 16.7
sd = 6.0
sigma = sd/sqrt(n)
true_mean = 15.5

a. State the null and alternative hypotheses. (3 points)

The Nulll Hypothesis can be stated as: The mean length of imprisonment for motor-vehicle-theft offenders in Australia is 16.7 months.

• $ H_0: \mu = 16.7 months. $

The alternative Hypothesis can be stated as: The mean length of imprisonment for motor-vehicle-theft offenders in Sydney differs from the national mean in Australia.

• $H_a: \mu\neq 16.7 months $

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b. Determine the probability of a Type I error. (2 points)

The probability of a Type I error is denoted by $ \alpha $, where $ \alpha $ is also known as the significance level of the hypothesis test. So here the probability of a Type I Error is: 0.05 or 5%.

This represents the probability of rejecting the null hypothessis when it is in fact TRUE.

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c. Determine the probability of a Type II error. (5 points)

The probability of a Type II error is denoted by $ \beta $. This represents the probability of NOT rejecting the null hypothesis when it is in fact false. To solve for $ \beta$, the critical values must be found.

Note: We have a 2-sided hyptohesis => 2 Critical Values.

# Solving For Critical Value 1
# folliwng slide 48
x_left = qnorm(alpha/2, mean = mu, sd = sigma, lower.tail = T)
x_left

15.524021609276

# Solving For Critical Value 2
# folliwng slide 48
x_right = qnorm(alpha/2, mean = mu, sd = sigma, lower.tail = F)
x_right

17.875978390724

# solving for Beta
# slide 49
curve1 = pnorm(x_right, mean = true_mean, sd = sigma, lower.tail = T)
curve2 = pnorm(x_left, mean = true_mean, sd = sigma, lower.tail = T)

beta = curve1 - curve2
beta

0.483994726023825

From above, we see that our probability of a Type II Error, $\beta$ is approx. 0.48 or 48%.

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d. Simulate 1,000 samples, each of size 100. (5 points)

Here we first created a normalized standard deviation of 100 datapoints with a mean of 15.5 and a standard deviation of 6.0 using rnorm. This is fed into the replicate function where we then create 1000 samples of the output of the rnorm function.
link to walkthrough we followed

set.seed(111)
# replicate function "repeatedly evaluate some expression in R a certain number of times."
samples = replicate(1000, rnorm(n, true_mean, sd))
head(samples,2)

16.91132	19.097718	13.70780	16.35879	13.173022	23.87493	5.026768	12.870508	13.88540	13.74923	...	4.075963	22.7285	13.12917	15.33994	11.75197	7.827524	15.71353	16.97241	17.33399	23.80199
13.51558	8.538023	11.93879	17.68144	6.992768	10.14808	20.238205	6.237921	31.25065	15.34757	...	9.393428	15.7630	26.93070	14.61639	11.88793	20.296309	11.09280	16.15424	10.96369	20.24181

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e. Determine the mean of each sample in part (d). (5 points)

Here we used the apply function to get the mean of every column/sample.
link

# margin = 2 --> applies to columns 
sample_means = apply(samples, 2, mean)
# get into matrix
sample_means = as.matrix(sample_means)
head(sample_means, 5)

15.42309

15.39880

16.31830

15.63875

14.33606

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f. For the 1,000 samples obtained in part (d), about how many would you expect to lead to nonrejection of the null hypothesis? Explain your answer. (5 points)

A non-rejection of the null hypothesis is equal to the probability of a Type II Error. Here our $\beta$ is 0.48 or 48%. So then, 48% of 1,000 is 480. We expect 480 samples to lead to nonrejection of the null hypothesis.

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g. For the 1,000 samples obtained in part (d), determine the number that lead to nonrejection of the null hypothesis. (5 points)

Here we need to find the samples that have a Sample Mean that is between our to critical values. I.e: $$ 15.52 < \bar{x} < 17.88 $$ To do so, we will first filter out the values that are above 15.52, and then filter those values again to get the values below 17.88.

above_xleft = sample_means[sample_means > x_left]

between = above_xleft[above_xleft < x_right]
length(between)

486

From above, we see that we have 486 samples that lead to nonrejection of the null hypothesis.

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h. Compare your answers from parts (f) and (g), and comment on any observed difference. (5 points)

Here our actual value was higher than our expected value by 6 samples. This can be due to some variations in calculations, randomness of data, etc.

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i. Plot the power curve for the range of true μ between 14 to 19. Interpret your plot. (5 points)

Power of a test can be defined mathematically as: $$1 - \beta $$ To do so, we will re-use our code from Part C. Since we know our critical values x_left, and x_right, will not change, we will only need to solve for our new values of $\beta$ as it is dependent on the true mean of the data.

true_mean = seq(14,19, by = 0.001)

curve1 = pnorm(x_right, mean = true_mean, sd = sigma, lower.tail = T)
curve2 = pnorm(x_left, mean = true_mean, sd = sigma, lower.tail = T)

beta = curve1 - curve2
power = 1 - beta

plot(true_mean, power,'l', xlab = expression(mu), 
     ylab = 'Power', main = 'Power Curve')