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.

Sample mean:

a<- c(65,77)
sample_mean <- mean(a)
sample_mean
## [1] 71

Margin of error:

ME <- (77 - 65) / 2
ME
## [1] 6

Standard Deviation :

n <- 25
z<-1.645
SD = ME * n^0.5 / z
SD
## [1] 18.23708

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?
sd <- 250                               
ME <- 25                                
z <- qnorm(0.95)                   
z
## [1] 1.644854

Sample size:

samp_size <- (((z * sd) / (ME))^2)
samp_size
## [1] 270.5543
  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.

Luke’s sample size should be larger as he wants to have the higher confidence interval. Higher confidence interval require higher z score, as well. (c) Calculate the minimum required sample size for Luke. z score:

sd <- 250                               
ME <- 25                                
zscore <- qnorm(0.995)                  
zscore
## [1] 2.575829

Sample size:

sample_size <- (((zscore * sd) / (ME))^2)
ceiling(sample_size)
## [1] 664

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? There is a slight differences in average reading and writing, it is close to a normal distribution with slightly right skewed with mean at 0.
  2. Are the reading and writing scores of each student independent of each other? The reading and writing scores of each student are independent, reading and writing test data will not affect each other.
  3. 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? H0: Difference between average reading and writing scores is zero. HA: Difference between average reading and writing scores is not zero
  4. Check the conditions required to complete this test. The histogram is close to normal and the conditions of Central Limit Theorem were met, which are:
  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?
X = -0.545
SD = 8.887
n = 200
Z = 1.96
ME = Z*SD/(n^0.5)

CI = X + c(-ME,ME)
CI
## [1] -1.7766754  0.6866754

As the CI includes 0, there is not sufficient evidence that the difference between scores is not 0. We cannot reject the null hypothesis; there is no significant difference between scores.

  1. What type of error might we have made? Explain what the error means in the context of the application. Type I error: Incorrectly rejecting the null hypothesis. Type II error: Incorrectly rejecting the alternative hypothesis. Since we did not reject the null hypothesis, we are at a risk of making a Type II error. In this moment, we should have noticed that we had convincing data that there is a difference in the reading and writing average scores but we did not find any evidence.

  2. 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. Since our results indicated that there is no difference in the reading and writing scores, we can expect that the confidence interval would include 0.

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.

X = 22.92 - 27.88
SE = (5.29^2/26 + 5.01^2/26)^0.5
Z = 2.326
ME = Z*SE

CI = X + c(-ME,ME)
CI
## [1] -8.283576 -1.636424

The 98% CI for the differences in fuel efficience between the two types of cars does not contain 0. We can determine that there is a significant differece in fuel efficiencies; manual is more efficient than automatic.

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%?

We need to find the Z-score that would give us a lower tail of 80%. For a large sample size per group, the Z-score for a lower tail of 80% would be about Z = 0.84.

Additionally, the rejection region extends 1.96 SE from the center of the null distribution for sigma= 0.5. This allows us to calculate the target distance between the center of the null and alternative distributions in terms of the standard error: 0.84SE + 1.96SE = 2.8*SE

X = 4
SD = 2.2
Z = 1.282
ME = 0.5


n = (Z * SD / ME) ^ 2
ceiling(n)
## [1] 32

They would need 32 enrollees.

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. H0: The average number of hours worked do not vary across groups. HA: The average number of hours worked do vary across groups

  2. Check conditions and describe any assumptions you must make to proceed with the test. The observations are independent within and across groups: true assuming no one works two of these jobs. The data within each group are nearly normal: true. The variability is similar across groups: true as assumption.

  3. Below is part of the output associated with this test. Fill in the empty cells.

total_mean <- 40.45
mean_all <- c(38.67, 39.6, 41.39, 42.55, 40.85)
SD_all <- c(15.81, 14.97, 18.1, 13.62, 15.51)
n_all <- c(121, 546, 97, 253, 155)
Groups <- 5
SSG = sum(n_all*(mean_all - total_mean)^2)
MSG = 501.54
SSE = 267382 
SST = SSE + SSG
MSE = SSE/1167
F = MSG/MSE
PrF = pf(F, 4, 1167, lower.tail = FALSE)
SSG
## [1] 2004.101
MSG
## [1] 501.54
SSE
## [1] 267382
SST
## [1] 269386.1
MSE
## [1] 229.1191
F
## [1] 2.188992
PrF
## [1] 0.06819325
  1. What is the conclusion of the test? Assuming a significance level of 0.05, we cannot reject the hypothesis as p > 0.05 (p = 0.068). Therefore, we conclude that there is no significant variation in work hours between groups.