A teacher wishes to study the amount of time students in his statistics course s
ID: 3223531 • Letter: A
Question
A teacher wishes to study the amount of time students in his statistics course spend each week in study for the course. He believes that the average should be the nominal 6 hours (two hours outside class for every hour in class). So he has the students keep track of and report the time spent in study during a typical week. A total of 9 students respond. The average time spent is 6.5 hours with a standard deviation of 2 hours. Is this enough evidence to show that the average time is not 6 hours of studying? A hospital wants to see how many gloves nurses use over the course of their shift, so the nurses keep track of their usage of gloves. Their results are given in the table below. Is there enough evidence to support the claim that this hospital uses significantly more gloves than the expected average of 50? Extra Credit: 2 samplings of the average test scores in a class are taken, and the results are shown below. Are they significantly different at alpha = .01 Class 1: 80, 100, 75, 60, 75, Class 2: 90, 85, 75, 80, 90, 75, 85, 60Explanation / Answer
10)
Conclusion: Since the calculated value is 0.196 is less than 2.306, hence we will accept null hypothesis and conclude that average time is equal to 6.
11)
Null hypo: mu = 50
Alternative Hypo: mu not equal to 50
SD = sqrt(10099.38 / 39) = 16.09
SE = 16.09 / sqrt(39) = 2.58
xbar = average = 57.375
t = (xbar - mu) / SE = (57.375 - 50) / 2.58
= 2.85
Critical value at n-1 or 40-1 =39 degrees of freedom at level of significance 0.05 is 2.021
Conclusion Since the calculated value is 2.85 greater than critical value 2.021, hence we reject null hypothesis and conclude that hospital uses significantly more gloves than the expected average of 50.
Given, = 6 Xbar = 6.6 n = 9 Standard Deviation (SD)= 9.17 Standard Error (S.E) = 3.057 Degrees of freedom(DF) = 8 t = ( Xbar - ) / SE t = 0.196 Level of significance % = 0.05 Critical Value =2.306Related Questions
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