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# Confidence Intervals

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4) In a survey conducted to determine cost of vacations, 164 individuals were randomly sampled. Each person was asked to assess the total cost of his or her most recent vacation. The average cost was \$1,386. Assuming that the standard deviation was \$400, find a 94% confidence interval for the average cost of all vacation trips. ANSWER- 1327.28,1444.72

5) Students in a stats class were polled by a surveyor attempting to establish a relationship between hours of study in the week immediately before a major exam and the marks received on the exam.

Hours of Study 25, 12, 18, 26, 19, 20, 23, 15, 22, 8

Exam Grades 93, 57, 55, 90,82,95,95,80,85,61

Sum of X=188, Sum of Y=793 Xsquared=3832 Ysquared=65,143 SumofXY=15,540

a) Draw a scatter diagram to examine the relationship.

b) Find the equation of the regression line to help predict the exam score on the basis of study hours. ANSWER Y=39.4007 + 2.1223 X

c) Calculate the correlation coefficient. What does the value tell you about the relationship between hours of study and exam score? ANSWER r=.7705 implies a positive linear relationship.

6) Consider the following random variable X which represents the various amounts of damage (in thousands of dollars) to a house during the year.

X 0 10 30 50

prob ? .05 .02 .001

a) What is the P(x=0)? ANSWER=.929

b) Find the expected value and the variance of damage amount? ANSWER 24.1775

c) What is the probability that the damage amount is 50 given that there is some damage to the house? ANSWER=.0148

a) If P(A)=1/3 and P(B to the Cpower) = 1/4 can A and B be mutually exclusive? Explain

b) Suppose the standard deviation of a data set is 2. If I add 5 to each value in this data set, what is the new standard deviation of this data set?

c) True or False: If the coefficient of determination is .2, then the proportion of variation in Y not explained by the variation in X is .8.

d) Find k (k>1)using the normal tables: Area between -1 and -k is .0137

##### Solution Summary

The confidence intervals of probabilities are examined for standard deviation. The relationship between hours of study in the work before a major exam and the marks received on the exams are provided.

##### Solution Preview

6) Since the probabilities should add to 1 , so Prob ( X= 0) =
1 - ( .05 +.02 + .001 ) = .929

Expected Value = Sum of ( X * P(x))
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