Critical Value Example
An official with an advocacy agency for the elderly believes that the average monthly cost of medication used to control cholesterol has increased. A national study shows that the average monthly cost for the last 5 years for cholesterol control medication is $72.58 per month. The official selects a random sample of 50 elderly people who use medication to control cholesterol and finds the average cost to be $73.45 with a standard deviation of $2.25. What type of hypothesis test should be performed? At a 0.05 level of significance, what is the critical value for the hypothesis test? At a 0.05 level of significance, what is the test value? The official should conclude that the average cost has increased -- true or false?[1]
Answer:
H0: µ ≤ 72.58
Ha: µ > 72.58 (claim)
n = 50, 
=
73.45, s = 2.25, α = 0.05
The claim is that it is increasing, so running the right-tail test through an appropriate statistics program (Excel, Minitab, etc.), we find that the test stat is 2.7341; the critical value is 1.645, the p-value is .0043
We reject the null hypothesis. There is enough evidence at the .05 level to support the claim.
Example
A
sample of 81 account balances of a credit company showed an average balance of
$1,200 with a standard deviation of $126.
a. Formulate the hypotheses that can be used to determine whether the mean of
all account balances is significantly different from $1,150.
b. Let α = .05. Using the critical value approach what is your conclusion?
Answer:
a. Since
we want to know if the mean is "significantly different" from $1,150,
the null hypothesis is that it is $1,150.
H0: μ = 1150
Ha: μ ≠ 1150
b. Since
we don't have the population standard deviation, use the t test statistic.
t = (xbar-μ0)/(s/√n)
= (1200-1150)/(126/√81)
= 50/14 = 3.57
The critical value for t for 80 degrees of freedom and &alpha/2=0.025 is
1.990.
Since the t-value=3.57 is greater than the critical value of 1.990, we reject H0
and conclude that the mean is significantly different from $1,150.[2]
