Answer:
See the explanation below.
Explanation:
n = sample size = 1000
X = proportion of adults that a will = 450 / 1,000 = 0.45
P-value = 0.5
a = level of significance = 5%, or 0.05
A. At the 5% significance level, can you conclude that the percentage of people who have a will is less than 50%?
Our hypotheses are:
Null hypothesis: u ≥ 0.5
Alternate hypothesis: u < 0.5
Since this indicates a left tailed test, the decision rule is to reject null hypothesis if p-value is less than the level of significance.
Z = - 0.05 / 0.0158113883008419 = - 3.1623
Using the excel function NORMSDIST(-3.1623), we have:
P- value = 0.000782701129001276, approximately 0.0008
Therefore, null hypothesis is rejected at 5% level of significance since the p-value 0.0008 is less than 5%, or 0.05.
Therefore, there is an enough reason to reject null hypothesis.
B. What is the Type I error in part A? What is the probability of making this error?
Since the probability is greater than 50%, the Type I error is therefore the statement in the question that less than 50% of adults have a will.
Also, the probability of making this Type I error is the 5% level of significance.
C. What would your decision be in part A if the probability of making a Type I error were zero?
The null hypothesis will be rejected if the probability of making a Type I error were zero in part A.