Hypothesis Tests and Confidence intervals for one and two samples. MEA NS Height
ID: 3326572 • Letter: H
Question
Hypothesis Tests and Confidence intervals for one and two samples. MEA NS Heights of tall buildings A researcher estimates that the average height of buildings of 30 or more stories in a large city is at least 700 feet high. A random sample of 10 buildings is selected, and the heights in feet are shown. At .05, is there enough evidence to reject the claim? 1. 485 511 841 725 615 520 535 635 616 82 (b) Find a 95% Confidence Interval for the true height of these buildings. (c) Find n if you want E to be 60. 2. Retention Test Scores A random sample of non-English majors at a selected college was used in a study to see if the student retained more from reading a 19th century novel or by watching it in DVD form. Each student was assigned one novel to read and a different one to watch, and then they were given a hundred point quiz on each novel. The test results ar the DVD scores? e shown. At 0.05, can it be concluded that the book scores are higher than Book 90 80 90 DVD 85 72 80 80 70 90 84 75 80 80 (b) Find a 95% Confidence interval for the true difference in the means. - un) Vin Use: df = n1 + n2-2 21-X2 t=-Explanation / Answer
Question 1
H0 : h >= 700 feet
Ha : h < 700 feet
Sample mean x = 606.5 Feet
standard deviation of sample s = 109.08 feet
Here standard error of sample mean se0 = s/ sqrt(n) = 109.08/ sqrt(10) = 34.494 feet
Test statistic
t = (x - )/ se0 = (606.5 - 700)/34.494 = -2.71
so for alpha = 0.05 and one tailed test and dF = 10 -1 = 9
tcritical = 1.833
so here l t l > t(critical) so we will reject the null hypothesis and can conclde that there is enough evidence to reject the claim.
95% confidence interval = x +- tcritical se0 = 606.5 + 1.833 * 34.494
= (543.27, 669.73)
Here E = 60
so tcritical * (109.08/sqrt(n)) = 60
so here n = 11
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