What should we conclude about the analysts claim? Solve the test statistic approach. Use the following statistics to solve the problem. n = 15 = $1831.067 S = $506.59

1a) Sx  is a known estimate: True or False

1b) Population mean is a known estimate:  True or False

1c) The objective of a confidence interval is not to bracket the fixed population mean:  True or False

1d) T-test discusses the deviation of a random variable  from its mean measured in standard deviation of the sample mean units:  True or False

1e) T-test is used to test a hypothesis about the population mean: True or False

Standard Normal Probability Distribution

2) Find the probability that the random value of Z is between +.65 and +.2.33.

T distribution is symmetric and t table gives values of t such that the probability of the larger t is equal to a given probability.

3a) P(t > t_o) = 0.10 given a sample size of 13.

Find t_o

3b) P(-t_o  > t > + t_o) = 0.05 given a sample size of 9.

Find t_o

3c) P(t <  2.1098) = ? given a sample size of 18.

Find the associated probability.

Consider the probability that a random interval x ± (t.05)S will contain the fixed μ (population mean) is .95  That is:

P( – t.05S ≤  μ ≤   + t.05S) = .95.

4) Construct a Confidence Interval given the following:

Sample size = 10,  = 131, S² = 971.11.

T test:

5) If a real estate market is strong, there will be a close relationship between the asking price for homes and the selling price.  Suppose that one analyst believes that the mean difference between asking price and selling price for homes in a particular market area is less than $2000.  To test this using an alpha level equal to .05, random sample of n=15 homes that have sold recently was selected.  The difference between asking price and selling price data from the sample is in the following:

What should we conclude about the analysts claim?  Solve the test statistic approach.  Use the following statistics to solve the problem.

n = 15     = $1831.067         S = $506.59