Question

.According to a survey, high school girls average 100 text messages daily (The Boston Globe, April 21, 2010). Assume the population standard deviation is 20 text messages. Suppose a random sample of 50 high school girls is taken.

a) what is the mean daily text messages of a sample of 50 high school girls?

b) what is the standard deviation of daily text messages of a sample of 50 high school girls?

c) what is the 90th percentile of daily text messages?

.A study by Allstate Insurance Co. finds that 82% of teenagers have used cell phones while driving (The Wall Street Journal, May 5, 2010). Suppose a random sample of 100 teen drivers is taken.

a) what is the standard deviation of proportion of 100 teen drivers who use cell phone while driving?

b) we are interested in the proportion of 100 teen drivers who use cell phone while driving. What type of distribution is it- Exponential distribution, t distribution, Uniform distribution, Normal distribution?

Which one is correct statement? Choose all applied.

	a.	If your population of interest is SUNY NP students, then mean GPA of SUNY NP students is a random variable
	b.	If your population of interest is SUNY NP students, then mean GPA of 100 selected SUNY NP students is a random variable.
	c.	If your population of interest is SUNY NP students, then mean GPA of 100 selected SUNY NP students is a sample statistics
	d.	If your population of interest is SUNY NP students, then mean GPA of SUNY NP students is a population parameter.

Which one is the correct statement? Choose all applied.

	a.	If you are interested in US households, mean income of 500 US households is a random variable.
	b.	If you are interested in US households, proportion of 500 US households who do not have any health insurance is a point estimate of proportion of all US households who do not have any health insurance.
	c.	If you are interested in US households, mean income of 500 US households is a point estimate of mean income of all US households.
	d.	If you are interested in US households, mean income of US households is a fixed number, that is, not a random variable.

Answer 1

Q1) Assume X=average number of text messages of the 50 randomly chosen high school girls.

a) Then E(X)=100 (as the population average is 100)

b) SD(X)=population SD/(square root of sample size)=20/squareroot of 50=2.828427

c)By Central limit theorem, as sample size 50 is more than 30, the distribution of X is approximately normal with mean 50 and SD 2.828427.

If b is the 90% quantile, then P(X<=b)=.90 or approximately, P(N(0,1)<=(b-100)/SD(X))=.90. Hence, using tables, we get  (b-100)/SD(X)=1.281552 or b=100+SD(X)*1.281552=100+2.828427*1.281552=103.6248

Thus 90% is 103.6248

Q2) a)The variance of the proportion is p*(1-p)/n, where n=sample size=100 and p=.82

Thus SD =square root of (.82*.18/100)=0.03841875

b) As the sample size is large enough (>30), we can use CLT and hence the distribution can be approximated by Normal.

Q3) b,c,d are TRUE options

Q4) a,b,c, d are all TRUE

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