Working backwards, Part II. (5.24, p. 203) A 90% confidence interval for a population mean is (65, 77). The population distribution is approximately normal and the population standard deviation is unknown. This confidence interval is based on a simple random sample of 25 observations. Calculate the sample mean, the margin of error, and the sample standard deviation.

low90 <- 65
hi90 <- 77
SM <- (hi90 + low90)/2

The sample mean is 71.

ME <- (hi90 - low90) / 2

The margin of error is 6.

n <- 25
df <- n -1
p <- 0.9
tVal <- qt((1-p)/2,df)
SE <- ME/tVal
SamSD <- SE * sqrt(n)
SamSD
## [1] -17.53481

The Sample standarddeviation is 17.5348146.

SAT scores. (7.14, p. 261) SAT scores of students at an Ivy League college are distributed with a standard deviation of 250 points. Two statistics students, Raina and Luke, want to estimate the average SAT score of students at this college as part of a class project. They want their margin of error to be no more than 25 points.

  1. Raina wants to use a 90% confidence interval. How large a sample should she collect?
SAT_sd <- 250
SAT_ME <- 25
SAT_z_90 <- 1.65

samples <- (((SAT_z_90 * SAT_sd)/(SAT_ME))^2)

Raina should use a sample size of at least `r ceiling(samples)’

  1. Luke wants to use a 99% confidence interval. Without calculating the actual sample size, determine whether his sample should be larger or smaller than Raina’s, and explain your reasoning.

Since Luke wants a higher confidence intrval, he would need a larger sample size.

  1. Calculate the minimum required sample size for Luke.
SAT_z_99 <- 2.58

samples_99 <- (((SAT_z_99 * SAT_sd)/(SAT_ME))^2)

The minimum sample size for a 99% confidence interval is 666

High School and Beyond, Part I. (7.20, p. 266) The National Center of Education Statistics conducted a survey of high school seniors, collecting test data on reading, writing, and several other subjects. Here we examine a simple random sample of 200 students from this survey. Side-by-side box plots of reading and writing scores as well as a histogram of the differences in scores are shown below.

  1. Is there a clear difference in the average reading and writing scores?

Though the actual means are different, the overlap is significant, so there isn’t necessarily a clear difference.

  1. Are the reading and writing scores of each student independent of each other?

Doubtful - a person’s reading ability and writing ability are probably related.

  1. Create hypotheses appropriate for the following research question: is there an evident difference in the average scores of students in the reading and writing exam?

Null: There is no difference in the average scores of students in reading and writing.

Alternate: There is a difference in the average scores of students in reading and writing.

  1. Check the conditions required to complete this test.

The two sets of data re not necessarily independent, but there is a large number of them.

  1. The average observed difference in scores is \({ \widehat { x } }_{ read-write }=-0.545\), and the standard deviation of the differences is 8.887 points. Do these data provide convincing evidence of a difference between the average scores on the two exams?
sd_diff <- 8.887
mean_diff <- -0.545
n <- 200

SE <- sd_diff / sqrt(n)

tVal <- (mean_diff)/ SE

pVal <- pt(tVal,n-1)
pVal
## [1] 0.1934182

With such a high p-value (0.1934182) we cannot reject the null hypothesis.

  1. What type of error might we have made? Explain what the error means in the context of the application.

As stated above, the two sets of data may not be independent, so our running this test may not be valid.

  1. Based on the results of this hypothesis test, would you expect a confidence interval for the average difference between the reading and writing scores to include 0? Explain your reasoning.

Yes, we would expect such a confidence interval to contain 0, since the test suggests there isn’t really a difference between the average reading and writing scores.

Fuel efficiency of manual and automatic cars, Part II. (7.28, p. 276) The table provides summary statistics on highway fuel economy of cars manufactured in 2012. Use these statistics to calculate a 98% confidence interval for the difference between average highway mileage of manual and automatic cars, and interpret this interval in the context of the data.

auto_mean <- 22.92
auto_sd <- 5.29
auto_n <- 26

man_mean <- 27.88
man_sd <- 5.01
man_n <- 26

z_98 <- 2.326

mean_diff <- auto_mean - man_mean

SE_diff <- sqrt((auto_sd^2 / auto_n) + (man_sd^2 / man_n))

low <- mean_diff - SE_diff * z_98
high <- mean_diff + SE_diff * z_98
low
## [1] -8.283576
high
## [1] -1.636424

Based on the fact that the confidence interval of (-8.2835765, -1.6364235) does not contain zero, there is ample evidence to state that there is a difference of fuel efficiency between automatic and manual vehicles.

Email outreach efforts. (7.34, p. 284) A medical research group is recruiting people to complete short surveys about their medical history. For example, one survey asks for information on a person’s family history in regards to cancer. Another survey asks about what topics were discussed during the person’s last visit to a hospital. So far, as people sign up, they complete an average of just 4 surveys, and the standard deviation of the number of surveys is about 2.2. The research group wants to try a new interface that they think will encourage new enrollees to complete more surveys, where they will randomize each enrollee to either get the new interface or the current interface. How many new enrollees do they need for each interface to detect an effect size of 0.5 surveys per enrollee, if the desired power level is 80%?

mean_surveys <- 4
sd_surveys <- 2.2
ME_surveys <- 0.5
z_80 <- 1.28

n_surveys <- (z_80 * sd_surveys / ME_surveys)

They would need a sample size of at least 6

Work hours and education. The General Social Survey collects data on demographics, education, and work, among many other characteristics of US residents.47 Using ANOVA, we can consider educational attainment levels for all 1,172 respondents at once. Below are the distributions of hours worked by educational attainment and relevant summary statistics that will be helpful in carrying out this analysis.

  1. Write hypotheses for evaluating whether the average number of hours worked varies across the five groups.

Null: The averages across all 5 groups are the same.

Alternate: The averages across some fo the 5 groups are different.

  1. Check conditions and describe any assumptions you must make to proceed with the test.

* We can assume independence.

* We can assume normalcy (as supported by the box plots).

* The variability across groups appears similar.

  1. Below is part of the output associated with this test. Fill in the empty cells.
means <- c(38.67, 39.6, 41.39, 42.55, 40.85)
SDs <- c(15.81, 14.97, 18.1, 13.62, 15.51)
Ns <- c(121, 546, 97, 253, 155)
df <- data.frame(means, SDs, Ns)

Tot_N <- sum(Ns)
DF <- 5 - 1
DF_res <- Tot_N - 5

PrF <- 0.0682
MeanSq <- 501.54
SumSq <- 267382

F_stat <- qf(1 - PrF, DF, DF_res)
MeanSqVar <- MeanSq / F_stat
SumProd <- DF * MeanSq

SumSqTot <- SumSq + SumProd
DFTot <- DF + DF_res
_ Df Sum Sq Mean Sq F-value Pr(>F)
degree 4 2006.16 501.54 2.1889312 0.0682
Residuals 1167 267,382 229.1255188
Total 1171 269388.2
  1. What is the conclusion of the test?

Due to the high p-value, we cannot reject the Null hypothesis and say definitively that there are differences in the means.