Purpose

In this project, students will demonstrate their understanding of the normal distribution, sampling distributions, and confidence intervals and hypothesis tests.

Question 1

Assume IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. If a person is randomly selected, find each of the requested probabilities. Here, x, denotes the IQ of the randomly selected person.
a. P(x > 140)

# Use the pnorm function to find probabilities in a normal distribution.

pnorm(140, mean = 100, sd = 15, lower.tail = FALSE)
## [1] 0.003830381
  1. P(x < 110)
pnorm(110, mean = 100, sd = 15, lower.tail = TRUE)
## [1] 0.7475075
  1. What is the probability that a random selected student will have an IQ between 80 and 120?
pnorm(120, mean = 100, sd = 15, lower.tail = TRUE) - 
pnorm( 80, mean = 100, sd = 15, lower.tail = TRUE)
## [1] 0.8175776

Question 2

Continue to assume IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. If a person is randomly selected, find each of the requested probabilities.
a. What is the probability of a randomly selected student will have an IQ greater than 110?

1 - pnorm(110, mean = 100, sd = 15, lower.tail = TRUE)
## [1] 0.2524925
  1. Suppose that a random sample of 12 students is selected. What is the probability that their mean IQ is greater than 110?
pnorm(110, mean = 100, sd = (15/sqrt(12)), lower.tail = FALSE)
## [1] 0.01046067

Question 3

Load and store the sample NSCC Student Dataset using the read.csv() function. Find the mean, standard deviation, and sample size of the PulseRate variable in this dataset. Do you think it is likely or unlikely that the population mean pulse rate for all NSCC students is exactly equal to that sample mean found?

nscc_students <- read.csv("/Users/Nlilly/downloads/nscc_student_data.csv")

 
PulseRateStudents <- subset(nscc_students, nscc_students$PulseRate != "NA")

#Find the mean pulse rate.
mean(PulseRateStudents$PulseRate)
## [1] 73.47368
## [1] 73.47368

#Calculate the standard deviation of the pulse rates.
sd(PulseRateStudents$PulseRate)
## [1] 12.51105
## [1] 12.51105

#The sample size is the numbers PulseRateStudents object., store it in the variable numberPulseRate.
numberPulseRate <- nrow(PulseRateStudents)
numberPulseRate
## [1] 38
## [1] 38

Question 4

Construct a 95% confidence interval for the mean pulse rate of all NSCC students and conclude your result in a complete sentence. (Note: we can create a valid confidence interval here since n > 30)

# Store mean and std dev

meanPulseRate <- mean(PulseRateStudents$PulseRate)
sdPulseRate <- sd(PulseRateStudents$PulseRate)

# Calculate lower bound of 95% CI

meanPulseRate - 1.96*(sdPulseRate/sqrt(numberPulseRate))
## [1] 69.49575
# Calculate upper bound of 95% CI

meanPulseRate + 1.96*(sdPulseRate/sqrt(numberPulseRate))
## [1] 77.45162

Question 5

Construct a 99% confidence interval for the mean pulse rate of all NSCC students and conclude your result in a complete sentence below the R chunk.

# Calculate lower bound of 99% CI
meanPulseRate - 2.58*(sdPulseRate/sqrt(numberPulseRate))
## [1] 68.23742
# Calculate upper bound of 99% CI
meanPulseRate + 2.58*(sdPulseRate/sqrt(numberPulseRate))
## [1] 78.70995

Question 6

Describe and explain the difference you observe in your confidence intervals for questions 4 and 5.

The 95% confidence interval is between (69.50, 77.45) which is about 8 bpm. The 99% confidence interval is between (68.23, 78.71) which is about 10.5 bpm. What expanding the confidence interval expands the range making the confidence interval higher because there is a better chance of being within the range.

Question 7

In the Fall 2009 semester of the 2009-10 academic year, the average NSCC student took 12.1 credits. I’m curious if that average differs among NSCC students last year (a sample of which is in the NSCC student dataset). Conduct a hypothesis test by a confidence interval to determine if the average credits differs last year from Fall 2009.

  1. Write the hypotheses (Try to emulate the “LaTeX” format I used in lecture notes. Otherwise, just give your best effort.)
    H0: μLastYear=μFall2009= 12.1 credits HA: μLastYear ≠ 12.1 credits

  2. Create confidence interval

# Store mean of Credits variable

meanCredits <- mean(nscc_students$Credits)

# Store standard deviation of Credits variable

sdCredits <- sd(nscc_students$Credits)


# Store sample size of Credits variable

numberCredits <- nrow(nscc_students)
# Lower bound of 95% CI
meanCredits - 1.96*(sdCredits/sqrt(numberCredits))
## [1] 10.73056
# Upper bound of 95% CI
meanCredits + 1.96*(sdCredits/sqrt(numberCredits))
## [1] 12.81944
  1. Make decision to reject \(H_0\) or fail to reject \(H_0\) based on confidence interval

We cannot reject the hypothesis based on the data.

  1. Write a concluding statement
    There is not enough evidence to prove that the average might be different.

Question 8

NSCC is investigating whether NSCC students have a higher than average stress level which can be identified by a higher than average standing pulse rate. Conduct a hypothesis test by a p-value to determine if NSCC students have a higher pulse rate than the national average of 72 bpm for adults.

  1. Write the hypotheses (Try to emulate the “LaTeX” format I used in lecture notes. Otherwise, just give your best effort.)

H0: μNSCC = μAdult= 72 bpm HA: μNSCC> 72 bpm

  1. Calculate p-value of getting sample statistics by chance
# Probability of getting sample data by random chance if mean was indeed 72bpm
pnorm(meanPulseRate, mean = 72, sd =(sdPulseRate/sqrt(numberPulseRate)), lower.tail = FALSE)
## [1] 0.2338856
  1. Make decision to reject \(H_0\) or fail to reject \(H_0\) at a significance level of 0.05 based on p-value.

We fail to reject the H0 because it is the p-value is greater than the significance level.

  1. Write a concluding statement

There is not enough sufficient evidence to support the claim that Norhtshore student’s have an above average stress level.