Chapter 5 - Inference for Numerical Data

Graded: 5.6, 5.14, 5.20, 5.32, 5.48

Q. 5.6 Working backwards, Part II. A 90% con???dence interval for a population mean is (65, 77). The population distribution is approximately normal and the population standard deviation is unknown. This con???dence 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:

df= 25-1 =24 two tails with 90% confidence: p= 0.05 => t=2.064

marigin of error = sample standard deviation/squart(n) * t = s/squart(25) * 1.711 = 0.3422s

65 = sample mean - marigin of error = x - 0.3422s

77 = sample mean + marigin of error = x + 0.3455s

=> sample mean: x= 71 margin of error: 0.3422s = 6 sample standard deviation: s= 17.53

Q. 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.

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

Ans: since the sample are indepandent, and the distribution of the error is center at 0, => use t-table.

standard deviation:s= 250 Max margin error = 25 = 250/squart(n) * t => 10= squart(n) / t 90% confidence interval => n = 272 (t=1.65 when n > 200)

  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.

Ans: Since twp tails becomes samller when df increase, the actual samole size needs to increse for Luke to get 99% confidence at the same boundary of the margin error as Raina’s.

  1. Calculate the minimum required sample size for Luke.

Ans: n=670 (t=2.59 when 500 < n < infi )

Q.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.

  1. Is there a clear difference in the average reading and writing scores? Ans: Yes.

  2. Are the reading and writing scores of each student independent of each other? Ans: Yes, if each student did reading and writing without cheating.

  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? Ans: H0: there is no difference in the average scores of students in the reading and writing exam H1: there is difference in the average scores of students in the reading and writing exam

or H0: x1 - x2 = 0 H1: x1 - x2 != 0

  1. Check the conditions required to complete this test. Ans: 1.Samples seem all indepent since it said the survey were from 200 random sample. 2.It is looking for the different from parised data reading and writting, and there is no need to check skewness since sample size is greater than 30.

  2. The average observed difference in scores is ¯ xreadwrite =0.545, and the standard deviationof the differences is 8.887 points. Do these data provide convincing evidence of a difference between the average scores on the two exams? Ans: SE=8.887/ squart(200) = 0.6284 T=(0.545-0)/0.6284 = 0.8673 df=199 since the p-value is greater than 0.2, so we faile to reject H0. There is difference in the average scores of students in the reading and writing exam.

  3. What type of error might we have made? Explain what the error means in the context of the application. Ans: Type II error, since we may have incorrectly failed to reject H0, but we are not able to detect it.

  4. 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, since we failed to reject H0.

Q. 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 eciency of cars with manual and automatic transmissions in terms of their average city mileage? Assume that conditions for inference are satis???ed.

Ans: First assume the conditions for infeerence are satisfied: 1. Each manual and automatic cars will be considered indepent. 2. Looking for the different of two sample means we can use t distribution.

Hypothesis test: H0: x1 - x2 = 0 H1: x1 - x2 != 0

SE= squart(3.58^2/26 +4.51^2/26) =squart(0.4929+0.7823) =1.1292 |x1 - x2| = | 16.12-19.85 |= 3.73 T= (3.73/1.1292)=3.3 df=26

since p-value is close to 0, we reject H0, these data provide strong evidence of a difference between the average fuel eciency of cars with manual and automatic transmissions in terms of their average city mileage

Q.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 ???ve groups. Ans: H0: the average of working hours is same for all attainment HA: the average of working hours is not same for all attainment

  2. Check conditions and describe any assumptions you must make to proceed with the test. Ans: 1.Indepent: assume each person is indepent to the others. 2.Approximate normal: samples size for each group is greater 30, we don’t need to check the skewness. 3.Constant variance: Variance in each group are close within (13.5,18.5).

  3. Below is part of the output associated with this test. Fill in the empty cells. df | sum sq | Mean Sq | F value | Pr(>F)

degree dfg SSG 501.54 MSG/MSE 0.0682 Residuals dfe 267382 MSE
Total 1171 269388.16

dfg = k-1 = 5-1 =4 dfe = n-k= 1172-5=1167 SSG = MSG * dfg = 501.54 * 4 = 2006.16 MSE = SSE/dfe = 267382/1167 = 229.1191 F = MSG/MSE = 2.189

  1. What is the conclusion of the test?

Since the p-value = 0.0682 which is greater than 0.05, we accept H0. The average of working hours is same for all attainment.