Question

#3 One of the most common ways of measuring intelligence is the IQ test. IQ scores in the US population have an average of µ = 100 and a standard deviation of σ = 15. Suppose a researcher wanted to test whether socioeconomic status (SES) has an effect on IQ scores. The researcher takes a random sample of n = 100 people, selected from a list of the 1000 richest people in the United States.

a. Based on this information, what is the expected value of the mean for the sampling distribution?

b. What is the standard error of the sampling distribution?

c. If my sample has an average IQ score of M = 99.25, what is the z score that represents the location of this value in the sampling distribution?

d. What is the probability of getting this result or greater? (Enter your response as a proportion in four decimal places.)

e. Assuming that z scores greater than +2 or less than -2 are unlikely to occur due to sampling error alone, is this a likely or unlikely result?

#8

A random sample of size 40 is to be selected from a population that has a mean μ = 48 and a standard deviation σ of 14. (Note: Most students will see a sample size of 30 or more, but if your question gives a sample size that is less than 30, you can assume that the population is normally distributed.)

This sample of 40 has a mean value of x, which belongs to a sampling distribution. Find the shape of this sampling distribution.

	a.	Approximately normal
	b.	Chi-square
	c.	Positively skewed
		d. Negatively skewed #9 a. Find the mean of this sampling distribution. (Give your answer correct to nearest whole number.) b. Find the standard error of this sampling distribution. (Give your answer correct to two decimal places.) c. What is the probability that this sample mean will be between 44 and 51? (Give your answer correct to four decimal places.) d. What is the probability that the sample mean will have a value greater than 53? (Give your answer correct to four decimal places.)

Answer 1

3)

a)  expected value of the mean for the sampling distribution=100

b) standard error of the sampling distribution =15/sqrt(100)=1.5

c) z score=(X-mean)/.std error =(99.25-100)/1.5 =-0.5

d)probability of getting this result or greater =P(Z>-0.5)=0.6915

e)it is likely event as does not fall outside of -2 and +2 region of usual events,

8)

option A) Approximately normal

a) mean of this sampling distribution =48

b) standard error of this sampling distribution =14/sqrt(40)=2.21

c)probability that this sample mean will be between 44 and 51

=P(44<X<51)=P((44-48)/2.21<Z<(51-48)/2.21)=P(-1.81<Z<1.36)=0.9131-0.0351 =0.8780

d)

probability that the sample mean will have a value greater than 53 =P(X>53)=P(Z>(53-48)/2.21)

=P(Z>2.26)=0.0119