HW5

5.6

Working backwards, Part II. 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 is in the middle of confidence interval.

mu = (77+65) / 2
mu

## [1] 71

margin of error is the difference of each of the bounds of confidence interval from the mean

ME <-(77-65)/2
ME

## [1] 6

sample standard deviation: sd = ME / T * sqrt(n)

T value for 90% confidence interval and df = 25-1 is 1.711 (from T-distribution table)

n=25
Tval = 1.711
sd <- ME / Tval * sqrt(n)
sd

## [1] 17.53361

5.48

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 are same across the five groups

H1: the average number of hours worked are different across the five groups

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

• The sample is assumed to be randomly selected and observations are independent

• Sample is less than 10% of the population and has greater than 30 observations

• Data should be normally distributed, box plot shows that almost all groups are normally distributed. Only “Bachelor’s” group shows skewness, but sample size is large enough to accept that fact.

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

Df column:

degree

K = 5
N = 1172
K-1

## [1] 4

residuals

N-K

## [1] 1167

total:

N-1

## [1] 1171

sum sq column:

SSdegree = MeanSqDegree*(K-1)

SSdegree = 501.54*(K-1)
SSdegree

## [1] 2006.16

SStotal = SSdegree + SSresiduals

SSresiduals = 267382
SStotal = SSdegree + SSresiduals
SStotal

## [1] 269388.2

mean sq column:

MeanSqResiduals = SSresiduals/(N-K)

MeanSqResiduals = SSresiduals/(N-K)
MeanSqResiduals

## [1] 229.1191

F-val = MeanSqDegree/MeanSqResiduals

MeanSqDegree = 501.54
F_val = MeanSqDegree/MeanSqResiduals
F_val

## [1] 2.188992

4. What is the conclusion of the test?

If we are using a p value < 0.05, then we fail to reject the null hypothesis and we can NOT conclude that the average number of hours worked are different across the five groups.