1. Suppose that the variable Z is distributed as Normal( = 0, = 1). Use Table A.
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Question
1. Suppose that the variable Z is distributed as Normal( = 0, = 1). Use Table A.3 to find the value of z1.
a. P(Z < z1) = 0.9505
b. P(Z > z1) = 0.0749
c. P(Z z1) = 0.0749
d. P(z1 < Z < z1) = 0.9500
2. New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204
per night (as of 2012). Assume that the room rates are normally distributed with a standard deviation of $55.
What is the cost of the top 20% most expensive hotel rooms in New York City?
3. A population has a mean of 200 and a standard deviation of 50. A sample size of 100 will be taken and the
sample mean x will be used to estimate the population mean.
a. What is the expected value of x (expected value is the average or mean value)?
b. What is the standard deviation of x?
c. Sketch the graph of the sampling distribution of x based on the sample size of 100?
d. What does the sampling distribution of x show?
4. Assume the population standard deviation is = 25.
a. Compute the standard error of the mean, x, for sample sizes of 50, 100 150, and 200.
b. What can you say about the size of the standard error of the mean as the sample size increases?
5. A simple random sample (SRS) of 60 items from a population of = 15 resulted in a sample mean of 80 (so,
x = 80).
a. Construct a 90% confidence interval for the population mean .
b. Construct a 95% confidence interval for the population mean .
c. Construct a 99% confidence interval for the population mean .
d. What happens to the size of the interval (max. – min.) as the confidence level increases from 90% to
99%?
6. A simple random sample from a normal population with n = 28 provided a sample mean x of 22.5 and a
sample standard deviation s of 4.4. {NOTE: You do not know the population standard deviation here,
so you must use the t distribution and determine the number of degrees of freedom (df).}
a. Construct a 90% confidence interval for the population mean .
b. Construct a 95% confidence interval for the population mean .
c. Construct a 99% confidence interval for the population mean .
7. The following sample data are from a normal population: 10, 8, 12, 15, 13, 11, 6, 5. The sample standard
deviation s = 3.46. {NOTE: You do not know the population standard deviation here, so you must use
the t distribution and determine the number of degrees of freedom (df).}
a. What is the point estimate of the population mean ?
b. What is the point estimate of the population standard deviation ?
c. With 95% confidence, what is the margin of error for the estimation of the population mean ?
d. What is the 95% confidence interval for the population mean ?
8. A simple random sample (SRS) from a normal population with n = 20 provided a sample mean x of 51.5 and
a sample standard deviation s of 6.84. {NOTE: You do not know the population standard deviation
here, so you must use the t distribution and determine the number of degrees of freedom (df).}
a. Construct a 95% confidence interval for the population mean .
b. Now assume this sample is twice as large. Instead of n = 20 use n = 40. Construct a 95% confidence
interval for the population mean .
c. What happens to the size of the interval (max. – min.) as the sample size increases from 20 to 40?
Explanation / Answer
1. The normal distribution table gives the value for z1 when you know the P(z<z1) and viceaversa
a. P(Z < z1) = 0.9505. look at the P value in the normal distribution table. The leftmost column gives a value of 1.6 and in the topmost row 0.05. so here z1 is 1.65
b. P(Z > z1) = 0.0749. . Since the normal distribution table gives value of P(z<z1). Thus, P(z<z1)=1-0.0748=0.9251 similar to the first one, z1 is 1.44
c. P(Z z1) = 0.0749. since it is a continuous distribution, the value remains same irrespective of the = sign. Thus answer is 1.44
d. P(z1 < Z < z1) = 0.9500. this is equal to 2*P(z<z1)-1=0.95
Thus this is P(z<z1)=1.95/2=0.945
Hence the z1 value is 1.6
Per chegg rules, I have answers the first question
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