Homework 1

For the problems in which calculations are needed, please include your R code, otherwise you will not be given full credit. Please upload your solution as an pdf or html file by Friday, September 5 at 11:59 pm under the assignments tab in Sakai.

1. Assume that \(X \sim \mathcal{N}(24,25)\).

• Calculate \(P(X \geq 10)\).

1 - pnorm(10, 24, 25)

[1] 0.7122603

• Determine the \(0.65-\)quantile of the standard normal random variable \(Z\).

qnorm(0.65, mean = 24, sd = 25)

[1] 33.63301

2. \(F_{a,b}\) is a random variable from the \(F\)-distribution with \(df1 = a\) and \(df2 = b\).

• Calculate \(P(F_{2,37} \leq 5.98)\)

pf(5.98, 2, 37)

[1] 0.9943809

• Find \(x\) such that \(P(F_{9,16} \geq x) = 0.2\)

qf(0.2, 9, 16)

[1] 0.5731567

#check
pf(0.5732, 9, 16)

[1] 0.2000329

3. Given a dataset of scores \(\{10,15,49,34,16,25,47,55,66,3,15,46,68\}\), calculate the mean, median, and variance. (Hint: the command c() creates a vector in R).

scores = c(10, 15, 49, 34, 16, 25, 47, 55, 66, 3, 15, 46, 68)
mean(scores)

[1] 34.53846

median(scores)

[1] 34

var(scores)

[1] 484.9359

4. A random sample of 40 persons attending a certain diet clinic was found to have lost an average of 18 pounds over a three week period, with a sample standard deviation of 6 pounds. For these data

• Calculate and interpret a 95% confidence interval for the true average weight loss.

xbar <- 18
s <- 6
n <- 40
alpha <- 0.05
crit <- qt(1-alpha/2, n-1)
CI <- xbar + c(-1, 1) * crit * s / sqrt(n)
CI

[1] 16.08111 19.91889

#We are 95% confident that the true mean weight loss (lbs.) over 3 weeks is between 16.08 and 19.92. 

• Suppose I wished to test my current belief that the average weight loss of the population is equal to 30 pounds. I come to the conclusion to fail to reject \(H_0: \mu = 15\) against the alternative \(H_a: \mu \neq 15\). Based on your answer to the first question, does this make sense? Why or why not?

No, with this confidence interval, there is evidence to reject H₀ , as 15 is not contained in the interval, and we cannot fail to reject H₀ if the value we’re testing against is not in the interval. The proper conclusion to make is to reject H₀ , as there is statistically significant evidence that the true mean weight loss is different than 15 lbs.

5. The dataset mice available directly in the R package datarium contains information about the weight, in grams, of 10 randomly selected lab mice. Suppose we believe that the average weight of this species of mice is 25 g. We would like to perform a t-test to determine if the average weight is different than 25 g.

• State the null and directional alternative hypothesis in symbols.

H₀ : 𝜇 = 25 g, H_a : 𝜇 ≠ 25 g

• Find the test statistic, degrees of freedom, and p-value for the data below. Be sure to clearly identify them from your output.

  library("datarium")
  test_stat = (mean(mice$weight) - 25)/(sd(mice$weight)/sqrt(length(mice$weight)))
  test_stat

  [1] -8.104504

  df_mice = length(mice$weight) - 1
  df_mice

  [1] 9

  t.test(mice$weight, mu = 25, sd = sd(mice$weight), n = length(mice$weight))

  
      One Sample t-test
  
  data:  mice$weight
  t = -8.1045, df = 9, p-value = 1.995e-05
  alternative hypothesis: true mean is not equal to 25
  95 percent confidence interval:
   18.78346 21.49654
  sample estimates:
  mean of x 
      20.14 

  p_val = 0.00001995
  p_val

  [1] 1.995e-05

• State your decision and conclusion for the given problem.

We are 95% sure at the ⍺ = 0.05 level that the true average weight of this species of mice is different than 25g since p < ⍺ (0.00001995 < 0.05)