Question

A consulting firm claims that the average age of millionaires in the US is under 55 years. To test this claim, a random sample of 30 millionaires was selected. The ages of those random millionaires are shown on the template.

40
51
60
25
30
55
76
61
38
65
78
54
50
74
80
27
67
42
46
26
55
40
78
48
46
33
31
80
57
49

A) state the appropraite null and alternative hypotheses.
B.)which test statistic do you use?

C.) what is the critical value of that test statistic, using a 0.01 significance leve?

D.) what is your decision rule?

E.) what is the calcualted value of the test statistic?

F.) do you reject or fail to reject the null hypothesis>

G.) what is your conclusion about the average age of millionares?

H.) calculate the p-value of this test and clearly state its meaning

Answer 1

Claim : The average age of millionaires in the US is under 55 years.

A) null and alternative hypotheses.

H₀ : µ = 55 vs H_a : µ < 55

We can find sample mean and sample standard deviation using excel function =AVERAGE( data set) and =STDEV.S( data set) respectively.

Therefore , $\bar{x}$ = 52.0667 and s = 17.1885

B) Population standard deviation σ is unknown therefore we use t test statistic.

C) Critical value:    

given α = 0.01

As H_a contain < sign , this is left tail test,therefore critical value would be negative.

d.f = n-1 = 30-1 =29

We can find critical value using excel function, =TINV(2*α ,d.f )

=TINV(2*0.01 ,29 ) = 2.462

Critical value = -2.462

D) Decision rule :

Reject H₀, if t test statistic ≤ -2.462 Or fail to reject H0 , if t test statistic > -2.462

E) Test statistic:

t = $\frac{\bar{X}-\mu }{S /\sqrt{n}}$ =  $\frac{52.0667-55 }{17.1885 /\sqrt{30}}$

t = -0.93

F) Decision: As t test statistic is greater than -2.462, we fail to reject H₀

G) Conclusion: There is no significant evidence that the average age of millionaires in the US is under 55 years.

H) To find p-value we can use excel function =TDIST( t , d.f , tail ) ( note : use positive value of t and for one tail test plug 1 for tail , for two tail test plug 2 for tail )

=TDIST(0.93,29,1)

P-value = 0.1800

Interpretation of p-value : The probability that sample mean  $\bar{x}$ is less than 52.0667 is equal to 0.1800, under the H₀ : µ = 55