Assignment: Hypothesis Testing in R

Problem 1: Student Sleep Analysis

The average time a student sleeps per night is summarized by the following data points:

6.5, 7.5, 9.0, 5.0, 6.5, 5.0, 6.0, 8.0, 4.5, 8.5, 6.5, 6.0

We will test the hypothesis:

• Null hypothesis \(H_0\): Students get an average of 8 hours of sleep per night (\(\mu = 8\)).
• Alternative hypothesis \(H_a\): Students get less than 8 hours of sleep per night (\(\mu < 8\)).

Significance level: \(\alpha = 0.05\)

(done together in class)

# Step 1: Store the data in a vector
sleeptime <- c(6.5, 7.5, 9.0, 5.0, 6.5, 5.0, 6.0, 8.0, 4.5, 8.5, 6.5, 6.0)


# Step 2: Calculate the sample mean and sample standard deviation
meansleep <- mean(sleeptime)
sdsleep <- sd(sleeptime)
meansleep 

[1] 6.583333

sdsleep

[1] 1.427543

#mean = 6.583333
#sd = 1.427543


# Step 3: Define the hypothesized mean and sample size
mu_0 <-8
n <- 12


# Step 4: Calculate the test statistic
test_statistic <- (meansleep-mu_0)/(sdsleep/sqrt(n))



# Step 5: Compute the p-value for a one-tailed test
p_value <- pt(test_statistic, n-1)


# Step 6: Make a conclusion based on p-value and significance level
#reject the null hypothesis and conclude that mean hours of sleep for students is less than 8

Problem 2

The average daily caffeine consumption (in mg) of a group of college students is recorded as follows:

400, 250, 0, 300, 0, 600, 250, 200, 0, 275

Perform a hypothesis test to determine if the students consume more than 260 mg of caffeine on average.

• State the null and alternative hypotheses.
• Use \(\alpha = 0.05\) as the significance level.
• Compute the test statistic, p-value, and make a conclusion.

Use the provided data and fill in the steps based on what you’ve learned from the first scenario.

#set the data to a vector named coffee
Coffee <- c(400, 250, 0, 300, 0, 600, 250, 200, 0, 275)
#determine mean, sample size, sd and null to find t stat
Cmean <- mean(Coffee)
sdc <- sd(Coffee)
csamplesize <- 10
cmu_0 <- 260
# determine t stat
t_stat <- (Cmean-cmu_0)/(sdc/sqrt(csamplesize))
#find p-value using tstat
cpvalue <- 1-pt(t_stat, csamplesize-1)
#conclusion based on p-value.69 fail to reject null hypothesis```

Problem 3

A two sample t-test is used for a hypothesis test that the mean of two groups are equal. If the two groups are equal size, the test statistic can be found using the formula: \[ t = \frac{\bar{x}_A - \bar{x}_B}{s_p \sqrt{\frac{2}{n}}} \] where \(s_p\) is the pooled standard deviation: \[ s_p = \sqrt{\frac{(n - 1)s_A^2 + (n - 1)s_B^2}{2n - 2}} \] \(\bar{x}_A\) and \(\bar{x}_B\) are the sample means for groups A and B respectively and \(s_A\) and \(s_B\) are sample standard deviations of the respective groups.

The table below provides the test scores of students in two groups:

Group A	Group B
85	88
87	84
83	89
86	87
84	85

For a two sample t-test, the null and alternative hypotheses are

• Null hypothesis \(H_0\): The mean scores of the two groups are equal (\(\mu_A = \mu_B\)).
• Alternative hypothesis \(H_a\): The mean scores of the two groups are not equal (\(\mu_A \neq \mu_B\)).

We will use significance level: \(\alpha = 0.05\)

Perform a two-sample balanced t-test to compare the means of the two groups. 4. Compute the p-value for a two-tailed test. 5. Compare the p-value with \(\alpha\) and make a conclusion.

#create Vectors for groups A and B 
groupA <- c(85, 87, 83, 86, 84)
groupB <- c(88, 84, 89, 87, 85)
#find each mean and sd for respective groups
meanA <- mean(groupA)
meanB <- mean(groupB)
sdA <- sd(groupA)
sdB <- sd(groupB)
#define sample size for each group
nA <- 5
nB <- 5
#calculate sd pooled
sp <- sqrt((nA-1)*(sdA^2)+(nB-1)*(sdB^2))/(nA + nB - 2)
#Calculate test statistic based on sp
group_tstat <- ((meanA)-(meanB))/(sp*(sqrt(2/5)))
#find p-value and make conclusion based on alpha level 0.05
group_pvalue <- 2*(1 - pt(group_tstat, sp))
#conclusion: fail to reject Null at p-value 1.74 and conclude that the means of the two groups are equal