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MATH 110 FINAL EXAM REVIEW

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1) The following graph and regression result gives the (hypothetical) IQ scores for 2 nd graders and the number of hours they watch TV each week. Simple linear regression results: Dependent Variable: IQ Independent Variable: TV hours/week IQ = 142.71914 - 3.1612654 TV hours/week Sample size: ...

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MATH 110 FINAL EXAM REVIEW 2023-2024
1) The following graph and regression result gives the (hypothetical)ndIQ
graders
scores for 2
and the number of hours they watch TV each week.



Simple linear regression results:

Dependent Variable: IQ
Independent Variable: TV hours/week
IQ = 142.71914 - 3.1612654 TV hours/week
Sample size: 20
R (correlation coefficient) = -0.86854228




nd
a) Find the predicted IQ for grader
a2 who watches 4 hours of TV per week.

Using the regression equation: IQ = 142.72 – 3.16 (4) = 130.08. The predicted IQ is
nd
approximately 130, for grader
a2 who watches 4 hours of TV per week.

b) Interpret the slope and y-intercept of the regression line in the context of the pro

Slope: The predicted IQ nd
ofgrader
a2 decreases by 3.16 as the number of TV hours per wee
increases by 1.

Y-intercept: The predicted IQndofgrader
a 2 who does not watch TV is approximately 142.7.



c) Interpret the correlation coefficient in the context of the problem.

The correlation coefficient is approximately -0.87. This tells us that is a fairly strong neg
linear association between IQ and TV viewing hours per week.

, 2) According to the Crime in the United States, 1998, 65% of murders are committed w
firearm. Suppose 500 murders are randomly selected.
Note: p=0.65 and n=500

a) What is the mean of the sampling proportion?

The mean of the sampling proportion is 0.65.


b) What is the standard deviation of the sampling proportion?


𝑝(1 − 𝑝)
𝑆𝐷 =
𝑛


0.65 (1 − 0.65)
≈ 0.0213
500

The standard deviation of the sampling proportion is 0.0213.


3) A company produces packets of soap powder labeled “Giant Size 32 Ounces.” The a
weight of soap powder in such a box has a Normal distribution with a mean of 33 oz
standard deviation of 0.7 oz. Suppose that 100 soap boxes are randomly selected.
Note: 𝜇 = 33, 𝜎 = 0.7, 𝑛 = 100

a) What is the mean of the sampling distribution?

The mean of the sampling distribution is 33 oz.

b) What is the standard deviation of the sampling distribution?

𝜎
𝑆𝐷 =
√𝑛
0.7
= 0.07
√100

The standard deviation of the sampling distribution is 0.07.

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