Question 1 – Sleep deprivation, CA vs. OR (2 pts)

A CDC report on sleep deprivation rates shows that the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents.

Goal: Conduct a hypothesis test to determine if these data provide strong evidence that the rate of sleep deprivation is different for the two states.

Part A: What type of hypothesis test is this? (proportion / diff. in proportions/ mean / diff. in means)

# diff in proportions

Part B: Compute $\widehat{p}_{pool}$.

(11545*.8 + 4691*.88) / (11545 + 4691)

## [1] 0.8231141

Part C: Compute the test-statistic Z.

(.88 - .8) / sqrt(.0823*(1-.0823)*(1/11545 + 1/4691))

## [1] 16.81241

Part D: Compute the p-value using pnorm().

2*pnorm(16.81, lower.tail=F)

## [1] 2.061714e-63

Part E: Write a conclusion to this question using ‘strength of evidence’.

There is overwhelmingly strong evidence that the two states have different rates of sleep deprivation. Looking at our sample statistics, it looks like Oregon has the higher rate.

Question 2 – Find the P-value with t-distribution (.5 pts)

A random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic using pt().

Part A: n = 26, T = 2.485, two-tail test

2*pt(2.485, df=25, lower.tail=F)

## [1] 0.0200048

Part B: n = 18, T = 0.5, right-tail test

pt(.5, df=17, lower.tail=F)

## [1] 0.3117426

Question 3 – Piano (1 pt)

Georgianna claims that in a small city renowned for its music school, the average child takes less than 5 years of piano lessons. We have a random sample of 35 children from the city, with a sample mean of 4.6 years of piano lessons and a sample standard deviation of 2.2 years.

Evaluate Georgianna’s claim using a hypothesis test.

state hypotheses
check conditions
calculate test-statistic
p-value
conclusion

Solutions:

$H_0$: $\mu$=5, $H_A$: $\mu <$ 5

There was a random sample of 35 $>$ 30 children, so the conditions are met to use a t-test.

# test-statistic
T = (4.6 - 5)/(2.2 / sqrt(35))
T

## [1] -1.075651

# p-value
pt(T, df=34)

## [1] 0.1448287

There is very little to no evidence to say that the mean years of piano lessons taken by children is below 5 years.

Question 4 – Diamonds (1.5 pts)

We have data on two random samples of diamonds: one with diamonds that weigh 0.99 carats and one with diamonds that weigh 1 carat. Each sample has 23 diamonds. Sample statistics for the price per carat of diamonds in each sample are provided below.

0.99 Carats

Sample mean = $44.51, s.d. = $13.32, n=23

1 Carat

Sample mean = $57.20, s.d. = $18.19, n=23

Assuming that the conditions for conducting inference using the t-distribution are satisfied, perform a hypothesis test to see if there is a difference in population prices per carat of diamonds that weigh 0.99 carats and 1 carat. (Wickham 2016)

Solutions:

# test-statistic
T = (57.2 - 44.51) / sqrt(13.32^2 / 23 + 18.19^2 / 23)
T

## [1] 2.699393

# p-value
pt(T, df=22, lower.tail=F)

## [1] 0.006548213