In: Statistics and Probability
A random sample of 145 recent donations at a certain blood bank
reveals that 81 were type A blood. Does this suggest that the
actual percentage of type A donations differs from 40%, the
percentage of the population having type A blood? Carry out a test
of the appropriate hypotheses using a significance level of
0.01.
State the appropriate null and alternative hypotheses.
H0: p = 0.40
Ha: p > 0.40
H0: p = 0.40
Ha: p <
0.40 H0: p =
0.40
Ha: p ≠ 0.40 H0:
p ≠ 0.40
Ha: p = 0.40
Calculate the test statistic and determine the P-value.
(Round your test statistic to two decimal places and your
P-value to four decimal places.)
z | = | |
P-value | = |
State the conclusion in the problem context.
Reject the null hypothesis. There is not sufficient evidence to conclude that the percentage of type A donations differs from 40%. Do not reject the null hypothesis. There is sufficient evidence to conclude that the percentage of type A donations differs from 40%. Reject the null hypothesis. There is sufficient evidence to conclude that the percentage of type A donations differs from 40%. Do not reject the null hypothesis. There is not sufficient evidence to conclude that the percentage of type A donations differs from 40%.
Would your conclusion have been different if a significance level
of 0.05 had been used?
Yes No
You may need to use the appropriate table in the Appendix of Tables
to answer this question.
Solution:
Here, we have to use one sample z test for the population proportion.
The null and alternative hypotheses for this test are given as below:
Null hypothesis: H0: The actual percentage of type A donations not differs from 40%.
Alternative hypothesis: Ha: The actual percentage of type A donations differs from 40%.
H0: p = 0.40
Ha: p ≠ 0.40
This is a two tailed test.
We are given
Level of significance = α = 0.01
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 = 81
n = sample size = 145
p̂ = x/n = 81/145 = 0.55862069
p = 0.40
q = 1 - p = 0.60
Z = (p̂ - p)/sqrt(pq/n)
Z = (0.55862069 – 0.40)/sqrt(0.40*0.60/145)
Z = 3.8989
Test statistic = Z = 3.90
P-value = 0.0001
(by using z-table)
P-value < α = 0.01
So, we reject the null hypothesis
There is sufficient evidence to conclude that the actual percentage of type A donations differs from 40%.
Reject the null hypothesis. There is sufficient evidence to conclude that the percentage of type A donations differs from 40%.
Would your conclusion have been different if a significance level of 0.05 had been used?
No
...because p-value is less than 0.05