Question 1 (30 points)

A medical researcher conjectures that the likelihood of having wrinkled skin around the eyes increases when a person smokes. The smoking habits as well as the presence of prominent wrinkles around the eyes were recorded for 500 randomly selected people from the population of interest. The following frequency table is obtained:

Prominent Wrinkles Wrinkles not prominent
Heavy smoker 95 55
Light smoker 75 75
Non-smoker 66 134
  1. Conduct a test to find out if someone’s smoking habits are associated with the presence of skin wrinkles. Use alpha= 0.05. You must follow these steps to conduct this test:

a1) State the hypotheses (Ho and Ha).

Ho: Skin wrinkles and smoking habits are independent.

Ha: Skin wrinkles depends on smoking habits

a2) Whether you reject or fail to reject Ho and why.

matrix_smoker_wrinkles = matrix(c(95,55,75,75,66,134), nrow=3, byrow=TRUE, dimnames= list(c("Heavy smoker","Light smoker","Non-smoker"),c("Prominent Wrinkles","Wrinkles not prominent")))

matrix_smoker_wrinkles
##              Prominent Wrinkles Wrinkles not prominent
## Heavy smoker                 95                     55
## Light smoker                 75                     75
## Non-smoker                   66                    134
chisq.test(matrix_smoker_wrinkles)$expected
##              Prominent Wrinkles Wrinkles not prominent
## Heavy smoker               70.8                   79.2
## Light smoker               70.8                   79.2
## Non-smoker                 94.4                  105.6
chisq.test(matrix_smoker_wrinkles, correct = FALSE)
## 
##  Pearson's Chi-squared test
## 
## data:  matrix_smoker_wrinkles
## X-squared = 32.32, df = 2, p-value = 9.59e-08

p-value = 9.59e-08

Since P-value is less than alpha (0.05), we reject Ho and support Ha,

a3) Your conclusion (i.e., whether smoking is associated with having wrinkles).

The data supports that skin wrinkles are caused by smoking

  1. Conduct a deeper analysis to know which smoking category is associated with prominent wrinkles, and which one is linked to non-prominent wrinkles. Show your work and discuss your results.
chisq.test(matrix_smoker_wrinkles, correct = FALSE)$residuals
##              Prominent Wrinkles Wrinkles not prominent
## Heavy smoker          2.8760653             -2.7192728
## Light smoker          0.4991518             -0.4719399
## Non-smoker           -2.9230237              2.7636713

The data shows that non-smokers are positive correlated with not prominent skin wrinkles, and Heavy and light smoker are positive correlated with prominent wrinkles. Also Heavy smokers are much related with prominent wrinkles than Light smokers.

Question 2 (40 points)

A researcher wants to compare the average anxiety levels of people living in Alaska and Hawaii. The researcher does not have any specific hypothesis in mind in terms of which state could have higher mean anxiety levels. She collected data on anxiety scores for two samples of randomly selected residents from both states. Each resident was given a score between 0 to 100 (higher scores mean more anxiety).

The anxiety scores she collected for each state are shown next. Create two vectors in R with these data (one vector for Alaska scores and another one for Hawaii scores).

Alaska scores: 69 76 64 65 67 77 56 67 62 82 56 77 71 68 76 69 64 66 83 77 75 79 71 75 86 67 70 73 77 71 78 64 62 58 67

Hawai scores: 64 76 74 74 73 71 75 63 67 77 74 67 69 70 64 72 72 72 74 76 67 69 80 73 68 77 71 73 69 68 71 71 73 75 71

Assume that the variables involved in this problem follow a normal distribution. Also assume their variances can be safely considered to be the same (in other words, you do NOT need to do the test to compare two variances here. Assume that the variances are equal).

alaska_scores = c(69, 76, 64, 65, 67, 77, 56, 67, 62, 82, 56, 77, 71, 68, 76, 69, 64, 66, 83, 77, 75, 79, 71, 75, 86, 67, 70, 73, 77, 71, 78, 64, 62, 58, 67)

hawaii_scores = c(64, 76, 74, 74, 73, 71, 75, 63, 67, 77, 74, 67, 69, 70, 64, 72, 72, 72, 74, 76, 67, 69, 80, 73, 68, 77, 71, 73, 69, 68, 71, 71, 73, 75, 71)
  1. State the hypotheses (Ho and Ha) that this researcher should set up to conduct the test that will allow her to make the desired comparison.

Ho: The level of anxiety in Alaska is less or equal than Hawaii

Ha: The level of anxiety in Alaska is greater than Hawaii

  1. Make a decision using a significance level of 0.05. Justify your decision.
t.test (alaska_scores, hawaii_scores, alternative = "greater", var.equal = TRUE)
## 
##  Two Sample t-test
## 
## data:  alaska_scores and hawaii_scores
## t = -0.70165, df = 68, p-value = 0.7574
## alternative hypothesis: true difference in means is greater than 0
## 95 percent confidence interval:
##  -3.376634       Inf
## sample estimates:
## mean of x mean of y 
##  70.42857  71.42857

Since P-value(0.7574) is greater than alpha(0.5), we cannot reject Ho. We conclude that the level of anxiety in Alaska is less or equal to Hawaii

  1. Obtain a 95% confidence interval for the difference between the mean anxiety level in Alaska and the mean anxiety level in Hawaii. Does the interval lead you to the same conclusion you reached from the hypothesis test? Justify.
t.test (alaska_scores, hawaii_scores, var.equal = TRUE, conf.level = 0.95)$conf.int
## [1] -3.843955  1.843955
## attr(,"conf.level")
## [1] 0.95

Since the CI doesn’t contain the zero, we can conclude that the average level of anxiety are different between Alaska and Hawaii.

Question 3 (15 points)

Consider a research study where the goal is to test whether nightly melatonin supplementation improves sleep (i.e., if it increases the amount of sleep time). The authors report the following result:

“In comparison with placebo, a 3 weeks of melatonin supplementation significantly increased the average sleep time (amount of increase= 36 min; P value = 0.046).”

Answer the following questions:

  1. What kind of hypothesis test was conducted by these authors? A test to compare two means? A test to compare two variances? A chi-square test to test for independence? Choose the correct option among these three and justify.

They conducted a two means test, because they were comparing the average sleeping time of people with a melatonin supplementation and a placebo group.

  1. State the hypotheses (Ho and Ha) that the authors were testing in this case. You need to state both Ho and Ha.

Ho: Melatonin supplementation doesn’t not improve sleeping time (less or equal sleep time)

Ha: Melatonin supplementation improves sleeping time (more sleep time)

  1. Did the authors find evidence to support the alternative hypothesis? Justify.

If they take an alpha of 0.05, p-value will be lower than alpha; therefore they could reject Ho and support Ha

Question 4 (15 points)

The results after rolling a die 300 times are shown in the next table:

1’s 2’s 3’s 4’s 5’s 6’s
Frequency 45 52 50 58 55 40

Is there sufficient evidence to conclude that a loaded die was used in this experiment? Justify by doing the relevant hypothesis test and showing all your work. You must state both hypotheses when doing the hypothesis test.

Note: A normal (not loaded) die is one with equal probability for all the faces of the die.

Ho: Probability of each number is 1/6

Ha: At least of one the probabilities is different from what we suspect/expect.

rolls = c(45, 52, 50, 58, 55, 40)

prob = rep(1/6,6)

Are all the expected counts greater than or equal to 5? Yes

chisq.test(rolls, p=prob)$expected
## [1] 50 50 50 50 50 50

Let’s run the test:

chisq.test(rolls, p=prob, correct= FALSE)
## 
##  Chi-squared test for given probabilities
## 
## data:  rolls
## X-squared = 4.36, df = 5, p-value = 0.4988

p-value = 0.4988

Since P-Value is greater than alpha(0.05) we cannot reject Ho, therefore the dice they used in the experiment was not loaded