In: Statistics and Probability
We are testing people to see if the rate of use of seat belts has changed from a previous value of 82%. Suppose that in our random sample of 75 people, we see that 66 have the seat belt fastened. The researcher is interested in α = 0.1 level test.
Step 1 State the null and alternative hypotheses.
Step 2 Write down the appropriate test statistic and the rejection region of your test (report z critical(s))
Step 3 Compute the value of the test statistic (z observed)
Step 4: State your conclusion (in one sentence, state whether or not the test rejects the null hypothesis and in another sentence apply the results to the problem).
Step 5: Compute the p- value for this test. State your decision.
Here, we have to use one sample z test for the population proportion.
Step 1 State the null and alternative hypotheses.
The null and alternative hypotheses for this test are given as below:
Null hypothesis: H0: the rate of use of seat belts is 82%.
Alternative hypothesis: Ha: the rate of use of seat belts is not 82%.
H0: p = 0.82 versus Ha: p ≠ 0.82
This is a two tailed test.
Step 2
We are given
Level of significance = α = 0.1
Critical value = - 1.6449 and 1.6449
(by using z-table)
Reject H0 if z < -1.6449 or z > 1.6449
Step 3
Test statistic formula for this test is given as below:
Z = (p̂ - p)/sqrt(pq/n)
Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size
x = number of items of interest = 66
n = sample size = 75
p̂ = x/n = 66/75 = 0.88
p = 0.82
q = 1 - p = 0.18
Z = (p̂ - p)/sqrt(pq/n)
Z = (0.88 - 0.82)/sqrt(0.82*0.18/75)
Z = 1.3525
Test statistic = 1.3525
Step 4:
Test statistic value lies between critical values, so we do not reject H0.
There is not sufficient evidence to conclude that the rate of use of seat belts has changed from a previous value of 82%.
Step 5:
P-value = 0.1762
(by using z-table)
P-value > α = 0.10
So, we do not reject the null hypothesis
There is not sufficient evidence to conclude that the rate of use of seat belts has changed from a previous value of 82%.