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 unknow. 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.

(77-65)/2=6, therefore the sample mean is 65+6=71. The z-score for 90% confidence interval is 1.645,
71+(1.645)(SE)/(√25)=77 Since ME=(z-score)(SE), We find 6=(1.645/5)SD, SD=18.237, SE=3.647

5.14 SAT scores. 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.

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

ME=25 SD=250 z-score for 90%=1.645 so

ME=(z-score)(SE)

25=(1.645)(250/(n^.5))

n=((1.645)(250/25))^2=270.6025, her sample should be 271.

(b) 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.

To be more confident, your z-score would increase. Therefore, the sample size would increase as well.

(c) Calculate the minimum required sample size for Luke.

n=((2.576)(250/25))^2=663.5776, so the minimum sample size would be 664.

5.20 High School and Beyond, Part I. 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.

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

There doesnt appear to be a difference in the average reading and writing scores.

(b) Are the reading and writing scores of each student independent of each other?

No, they are not.

(c) 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?

Ho: The difference of the average scores of students in the reading and writing exams is zero. H1: The difference of the average scores of students in the reading and writing exams is not zero. #(d) Check the conditions required to complete this test. n=200>30 simple random sample, relatively normal distribution

(e) The average observed difference in scores is ¯xreadwrite = 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?

n=(200)^(.5)
(.545-0)/(8.887/n)=.867, so the p-value is .192971
Therefore we say these data don't provide convincing evidence of a difference between the average scores on the two exams.

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

Type II error. We could have failed to reject the null hypothesis aka we could have said there is no difference between the test scores when in actuality there was.

(g) 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, when we assumed 0 was the mean our observed differnce was .545 and had a 19% chance of occuring, so I would assume 0 to be in our interval. Say we had a 95% CI, then our interval would be (.545-(1.96)(8.887/(200)^(.5)), .545+(1.96)(8.887/(200)^(.5))) = (-.6867, 1.7767), this interval includes 0.

5.32 Fuel eciency of manual and automatic cars, Part I. Each year the US Environmental Protection Agency (EPA) releases fuel economy data on cars manufactured in that year. Below are summary statistics on fuel eciency (in miles/gallon) from random samples of cars with manual and automatic transmissions manufactured in 2012. Do these data provide strong evidence of a difference between the average fuel efficiency of cars with manual and automatic transmissions in terms of their average city mileage? Assume that conditions for inference are satisfied.

Ho: There is no difference between the average fuel efficiency of cars with manual and automatic transmissions. H1: There is a difference between the average fuel efficiency of cars with manual and automatic transmissions. α <- 0.05 md<- 19.85-16.12=3.73

se <- (((3.58)^2)/26 + ((4.51)^2)/26)(.5)=1.129

t <- (3.73-0)/se = 3.304 df=25 p <- P(|t| >= 3.304)*2= .002877 Since p<.05, we reject the null hypothesis. This suggests that there is indeed a difference between the average fuel efficiency of cars with manual and automatic transmissions.

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

Ho: mlhs=mhs=mjc=mb=mg H1: they are not equal

(b) Check conditions and describe any assumptions you must make to proceed with the test.

all samples are well above 30, observations are independent, and the disributions are nearly normal (except bachelors, but the sample size is large enough to ignore the skew)

(c) Below is part of the output associated with this test. Fill in the empty cells.

a <-  c(38.67, 15.81, 121) 
b <- c(39.6, 14.97, 546) 
c <- c(41.39, 18.1, 97) 
d <- c(42.55, 13.62, 253) 
e <- c(40.85, 15.51, 155)
table <- data.frame(cbind(a, b, c, d, e))
stacked <- stack(table)
anova <- aov(values~ind, data = stacked)
summary(anova)

##             Df Sum Sq Mean Sq F value Pr(>F)
## ind          4  44648   11162   0.476  0.753
## Residuals   10 234275   23428

What is the conclusion of the test? We fail to reject the null hypothesis since Pr(>F) > .05, so there is not enough evidence to say there is a difference between the means.

Data606 ch 5