1. A study of commuting times reports the travel times to work of a random sampl
ID: 3361908 • Letter: 1
Question
1. A study of commuting times reports the travel times to work of a random sample of 30 employed adults in the Bay Area. The sample mean is 31.25 minutes and the standard deviation is 23.88 minutes. Find the 95% confidence interval to estimate the mean commute time for all employed adults in the Bay Area and write a sentence to interpret the interval. 2. An observational study was designed to compare the grades (in percent of total points earned) of students who sit in the front of a classroom with students who sit in the back. The following results were obtained: Sample standard deviation 9.36 12.77 n Sample mean rnt8 85.25 Back 5 69.20 ) Peciom alpsonuen e el o front of a classroom earn higher grades on average. If an error has been committed, which type of error is it? Explain the consequences of the error in the context of the problem. b) Find the 90% confidence interval estimate for the difference between the mean grade of students who sit in the front and the mean grade of students who sit in the back and write a sentence to interpret the interval. c)Explanation / Answer
Question 1
sample size n = 30
sample mean x = 31.25 minutes
sample standard deviation s = 23.88 minutes
95% confidence interval = x +- tdf, a/2 (s/ n)
95% confidence interval to estimate the mean commute time for all employed adults = 31.25 +- 2.0423 * (23.88/ 30)
= 31.25 +- 2.0423 * 4.36
= (22.346, 40.154)
Question 2
(a) Here,
H0 : back = front
Ha : back < front
Here
Pooled standard deviation sp = sqrt [{(n1 -1)s12 + (n2 -1)s22 }/(n1 + n2 -2)}]
sp = sqrt [(7 * 9.362 + 4 * 12.772)/11] = 10.7262
standard error of differnece = sp * sqrt [1/n1 + 1/n2] = 10.7262 * sqrt [1/8 + 1/5] = 6.115
Test statistic
t = (85.25 - 69.20)/ 6.115 = 2.6247
so at alpha = 0.05 , tcritical = t11,0.05 = 1.7959
so t > tcritical so we shall reject the null hypothesis and can claim that students who sit in the front of classromm earn hgher grades on average.
(B) Here if an eerror has been committed, the ype of error is ype I error. Here the consequences of this error is that we will reject the null hypothesis even if it is correct. That means we will say that we will conclude that there is effect of position of sitting in class even if it is not present here.
(C) 90% confidence inerval = (xfront - xback) +- tcritical sp * sqrt (1/n1 + 1/n2 )
= (85.25 - 69.20) +- 1.7959 * 10.7262 * sqrt [1/8 + 1/5]
= 16.05 - 10.982
= (5.068,27.032)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.