Title: HW Chapter 7 - Inference for Numerical Data

Author: John Mazon

Class: Data606

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

n <- 25
x1 <- 65
x2 <- 77

SMean <- (x2 + x1) / 2
##Sample mean

n <- 25
x1 <- 65
x2 <- 77

ME <- (x2 - x1) / 2
##Standard Deviation

df <- 25 - 1
p <- 0.9
p_2tails <- p + (1 - p)/2

t_val <- qt(p_2tails, df)

# Since ME = t * SE
SE <- ME / t_val

# Since SE = sd/sqrt(n)
sd <- SE * sqrt(n)

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?
z <- 1.65 # due to 90% Confidence interval
ME <- 25
sd <- 250

n <- ((z * sd) / ME ) ^ 2
  1. Luke wants to use a 99% confidence interval. Without calculating the actual sample size, determine whether his sample Luke’s sample should be larger since it will require a higher z number multiplied by the standard deviation and then squared.

should be larger or smaller than Raina’s, and explain your reasoning. (c) Calculate the minimum required sample size for Luke.

z <- 2.575 # due to 99% Confidence interval
ME <- 25
sd <- 250

n <- ((z * sd) / ME ) ^ 2

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? I do not see a clear difference in the average of the reading and writing scores. The difference distribution is fairly normal around the zero difference, though it seems to be a slight skew to the right.
  2. Are the reading and writing scores of each student independent of each other? I would say that the scores are independent of each student but not of each score, that is reading and writing scores are not independent of each other for each student.
  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? H_0: The difference of average in between reading and writing equal zero. That is: μr−μw=0

H_A: The difference of average in between reading and writing does NOT equal zero. That is: μr−μw≠0 (d) Check the conditions required to complete this test.

Independence of observations: The difference histogram suggested the data are paired. If paired, then they wouldn’t be independent.

Observations come from nearly normal distribution: The box plot provided in the text suggests the data are reasonably normally distributed and no outliers exist.
  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?

H0: The difference of average scores is equal to zero. That is: μdiff=0

HA
The difference of average scores is NOT equal to zero. That is: μdiff≠0

The paired data is presumably from less than 10% of the population of senior high schoolers, and from a simple random sample. We noted that the differences are nearly normally distributed, so the conditions are met in order to apply the t-distribution.

  1. What type of error might we have made? Explain what the error means in the context of the application. Type I error: Incorrectly reject the null hypothesis.

Type II error: Incorrectly reject the alternative hypothesis.

  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.

Certainly, I would expect a confidence interval for the average difference between reading and writing scores to include 0.

When the confidence interval include 0 for this kind of hypothesis test, it indicates that the difference is not in one side or another.

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. H0: The difference of average miles is equal to zero. That is: μdiff=0

HA
The difference of average miles is NOT equal to zero. That is: μdiff≠0 

n <- 26
# Automatic
mu_a <- 16.12
sd_a <- 3.58
# Manual
mu_m <- 19.85
sd_m <- 4.51

# difference in sample means
mu_Diff <- mu_a - mu_m

# standard error of this point estimate
SE_Diff <- ( (sd_a ^ 2 / n) + ( sd_m ^ 2 / n) ) ^ 0.5

t_val <- (mu_Diff - 0) / SE_Diff
df <- n - 1
p <- pt(t_val, df = df)
p
## [1] 0.001441807

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%?

Assuming significance level is α = 0.05, 304 new enrollees are needed for each interface to detect an effect size of 0.5 surveys per enrollee, if the desired power level is 80%.

sd1 <- 2.2
sd2 <- 2.2
effect_size <- 0.5
standard_error <- 0.5/(0.84+1.96)
n <- (sd1^2+sd2^2)/standard_error^2
n
## [1] 303.5648

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 doesn’t vary accross the five groups.

H1: The average number of hours worked varies accross the five groups.

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

    The cases are independent from each other.

    The distribution of the data is nearly normal.

    The deviation across groups are similar.

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

k <- 5
df_g <- k-1
df_g
## [1] 4
n <- 1172
df_e <- n-k
df_e
## [1] 1167
df_t <- df_g + df_e
df_t
## [1] 1171
MSG <- 501.54
SSG <- MSG*df_g
SSG
## [1] 2006.16
SSE <- 267382
SST <- SSE+SSG
SST
## [1] 269388.2
MSE <- SSE/df_e
MSE
## [1] 229.1191
f<-MSG/MSE
f
## [1] 2.188992
  1. What is the conclusion of the test? The P-value is 0.0682 > 0.05, we don’t reject the null hypothesis. The average number of hours worked doesn’t vary accross the five groups.