Chapter 7 - Inference for Numerical Data

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.

Ans:

Sample mean = 71; margin of error = 6; sample standard deviation = 17.53481.

n <- 25
df <- n-1
sample_mean <- (65+77)/2
sample_mean

## [1] 71

MOE <- (77-65)/2
MOE

## [1] 6

t <- qt(.95, df)
SE <- MOE/ t
sd <- SE*sqrt(n)  #sd/sqrt(n) = SE
sd

## [1] 17.53481

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.

Raina wants to use a 90% confidence interval. How large a sample should she collect?

Ans:

She should collect 271 samples.
```
sd <- 250
MOE <- 25
z <- qnorm(0.95)  # z=1.645
SE <- MOE/z
n <- (sd/SE)^2
n
```
```
## [1] 270.5543
```
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.

Ans:

Luke’s sample should be larger than Raina’s if he wants to use 99% confidence interval.
```
# As sample size n = (sd/SE)^2 = (sd *z /MOE)^2, 
# when z increases, n should be larger to keep MOE<=25.
```
Calculate the minimum required sample size for Luke.

Ans:

The minimum sample size required for Luke is 664.
```
(sd*qnorm(.995)/25)^2
```
```
## [1] 663.4897
```

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.

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

Ans:

There is no clear difference in the average reading and writing scores based on the plots.
Are the reading and writing scores of each student independent of each other?

Ans:

No, the reading and writing scores are dependent of each other as they are a pair of scores come from the same student.
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?

Ans:

H0: There is no difference in the average scores of students in the reading and writing exams.

H1: There is evident difference in the average scores of students in the reading and writing exams.
Check the conditions required to complete this test.

Ans:
1. Each pair of scores is independent from the other pairs, which, the scores of different students must be independent.
2. We assume the population is large that the sample size 200 is less than 10% of the population.
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?

Ans:

As we are using two-tailed test, our p-value = 0.3858 >0.05, we accept the null hypothesis. There is no difference in the average scores of students in the reading and writing exams.
```
n <- 200
df <- n-1
diff <- -0.545
sd <- 8.887
SE <- sd/sqrt(n)
z <- (diff-0)/SE
z
```
```
## [1] -0.867274
```
```
p <- pnorm(z)
p*2
```
```
## [1] 0.3857919
```
What type of error might we have made? Explain what the error means in the context of the application.

Ans:

We might have made Type 2 Error by failing to reject H0. It means that there is evident difference in the average scores of students in the reading and writing exams but we failed to prove so.
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.

Ans:

Yes, based on the results, we believe there is no (evident) difference in the average scores of students in the reading and writing exams, which means the difference of their means might possibly be 0. Therefore, we would expect a confidence interval to include 0.
```
# 95% confidence interval
c(diff-1.96*SE, diff+1.96*SE)
```
```
## [1] -1.7766754  0.6866754
```

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.

Ans:

By using the data provided in the table, we have the 98% confidence interval (-8.511, -1.409).

# both n for sample automatic and sample manual are 26
n <- 26
# degree of freedom
df <- n-1

# means_diff = mean_1 - mean_2
means_diff <- 22.92-27.88
means_diff

## [1] -4.96

# SE_(mean_x1 - mean_x2) = sqrt( (s_1)^2/n_1 + (s_2)^2/n_2 )
SE_diff <- sqrt(5.29^2/26 + 5.01^2/26)
SE_diff

## [1] 1.428881

# t-value at 98% ci with df=25
t_df <- qt(.99, n-1)
t_df

## [1] 2.485107

# margin of error of the difference
MOE_diff <- t_df * SE_diff
MOE_diff

## [1] 3.550922

# 98% confidence interval
c(means_diff - MOE_diff, means_diff + MOE_diff)

## [1] -8.510922 -1.409078

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%?

Ans:

Assume all surveys completed by enrollees are independent and the population is large.

By the calculation below, they need 304 new enrollees for each interface to detect an effect size of 0.5 surveys per enrollee with desired power level at 80%.

s1 <- 2.2
s2 <- 2.2

# minimum effect size
e <- 0.5

# critical value when alpha = 0.05
z1 <- qnorm(0.975)
z1

## [1] 1.959964

# critical value of 80% confidence interval
z2 <- qnorm(0.80)
z2

## [1] 0.8416212

# Sample size determination: effect size = (z1+z2)*SE = (z1+z2)*(s1^2/n + s2^2/n)
n <- (z1 +z2)^2 * (s1^2 + s2^2) / e^2
n

## [1] 303.9086

Work hours and education. (7.42, p. 297)

The General Social Survey collects data on demographics, education, and work, among many other characteristics of US residents. 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.

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

Ans:

H0: The average number of hours worked of all five groups are the same.

H1: At least one of the average number of hours worked among the five groups is different.
Check conditions and describe any assumptions you must make to proceed with the test.

Ans:
1. Assume the observations are independent within and across groups.
2. Assume the data within each group are nearly normal.
3. Assume the variability across the groups is about equal.
As the sample sizes are all larger than 30, and is less than 10% of the populaton, the conditions are met.
Below is part of the output associated with this test. Fill in the empty cells.

Ans:

# number of groups = k = 5
k <- 5
df_G <- k-1
df_G

## [1] 4

# number of total observations = n = 1172
n <- 1172
df_E <- n-k
df_E

## [1] 1167

# df_total = df_G + df_E
df_total = df_G + df_E
df_total

## [1] 1171

# Mean_sq_degree given = 501.54
MSG <- 501.54

# Sum_Sq_degree = Mean_Sq_degree * df_G
SSG <- MSG * df_G
SSG

## [1] 2006.16

# Sum_sq_residuals given = 267382
SSE <- 267382.00

# Sum_sq_total = sum_sq_degree + Sum_sq_residuals
SST <- SSG + SSE
sprintf("%f", SST)

## [1] "269388.160000"

# Mean_sq_residuals = Sum_sq_residuals / df_E
MSE <- SSE / df_E
MSE

## [1] 229.1191

# F-statistics
f <- MSG / MSE
f

## [1] 2.188992

(table)	Df	Sum Sq	Mean Sq	F-value	Pr(>F)
degree	4	2006.16	501.54	2.188992	0.0682
residuals	1167	267,382	229.11911
Total	1171	269388.16

What is the conclusion of the test?

Ans:

As the p-value = 0.0682 > 0.05, we fail to reject the null hypothesis.

There is no significant difference between the average number of hours worked of all five groups. Any observed difference is due to chance.

Chapter 7 - Inference for Numerical Data

Sin Ying Wong

10/27/2019

Working backwards, Part II. (5.24, p. 203)

SAT scores. (7.14, p. 261)

High School and Beyond, Part I. (7.20, p. 266)

Fuel efficiency of manual and automatic cars, Part II. (7.28, p. 276)

Email outreach efforts. (7.34, p. 284)

Work hours and education. (7.42, p. 297)