Question

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Q1: Suppose you survey that a random sample of 150 people from one hospital has been given cholesterol tests, and 60 of these people had levels over the “safe” count of 200. Construct a 95% confidence interval for the population proportion of people in test with cholesterol levels over 200.

Q2: A research conducted by the University of Michigan claimed that there are more female drivers in the USA than male drivers. A researcher decides to test this claim on his state. In his simple random sample of 900 observations, he noticed that 468 of 900 were women. At  = 0.05, is there enough evidence to support the claim? (Use the P-value method where (? < 1.20) = 0.8849)

Q3:

A sample of 100 body temperatures has a mean of 98.8℉. Assume that ? is known to be 0.6℉. Use a 0.05 significance level to test the claim that the mean body temperature of the population is equal to 98.6℉, as is commonly believed. Is there sufficient evidence to conclude that the common belief is wrong?

(Use the P-value method where (? < 3.33) = 0.9996)

Q4:

Listed below are the body lengths (in inches) and weights (in lb) of randomly selected bears:

Length

40      64        65        49        47

Weight

65      356      316      94        86

            •           Find the value of the linear correlation coefficient.

            •           Letting y represent weights of bears and letting x represent their lengths, find the regression equation.

            •           Based on the given sample data, what is the best predicted weight of a bear with a length of 72.0 inch?

Answer 1

Question 1)

Answer)

Sample size N = 150

Point estimate P = 60/150

First we need to check the conditions of normality, that is if n*p and n*(1-p) both are greater than 5

N*p = 60

N*(1-p) = 90

As both are greater than 5, conditions are met, so we can use standard normal z table to construct the interval

From z table, critical value z for 95% confidence level is 1.96

Margin of error(MOE) = z*√p*(1-p)/√n

N = 150

P = 60/150

Z = 1.96

After substitution

MOE = 0.0784

Confidence interval is given by

P-MOE < P < P+MOE

0.3216< P < 0.4784