Question

4. The following table gives the age (in years) of 36 randomly selected U.S. millionaires. The sample mean x−=58.53 years. Assume that the standard deviation of ages of all U.S. millionaires is 13.0 years.  

1. Test at the 10% significance level whether the average year of all U.S. millionaires is above 55 years.

2. What is the P-value of the hypothesis test in part (a)?

3. Obtain a confidence interval corresponding to the hypothesis test in part (a).

4. Does the interval in part (c) support the result in part (a)? Explain your answer.

Answer 1

#Solution:

parta

Given:

n=36

xbar=58.53

=13

=55

AnsaL:

Null and alternative hypothesis are

Ho:=55

Ha:>55

test statistics

here we use z-test as is known and sample size is large

Z=(xbar-)//sqrt(n)

=(58.53-55)/13/sqrt(36)

Z=1.629

#P-value=P(Z>1.629)

=0.0516

P-value=0.0516

hence p-value<0.10 hence we reject Ho

Conclusion: average year of all U.S. millionaires is above 55 years

partb:

#P-value=0.0516

partc:

90% confidence interval for

(xbarZ*/sqrt(n))

Z/2 at 90% CI is Z0.10/2=1.645

=58.53+/-1.645*(13/sqrt(36))

=58.53+/-3.56

CI=(54.97, 62.09)

partd:

from c and a

At 90% confident level we calulate the confidence interval can interpretate that true population mean is content (54.97,62.09)

the interval support the hypothesis test outcome that population mean is greater than 55