Answer:
Explanation:
Given:
Sample size, n= 1000
![X = (450)/(1000) = 0.45](https://img.qammunity.org/2021/formulas/mathematics/college/zri575yjkpym5v7z3t63g2amx280qbktvm.png)
P' = 0.5
Level of significance, a = 0.05
a) The null and alternative hypotheses:
H0 : u ≥ 0.5
H1 : u < 0.5
This is a left tailed test.
Decision:
Reject null hypothesis H0 if p value is less than level of significance.
The test statistic, z =
z = - 3.1623
From the standard normal table,
NORMSDIST(-3.1623)= 0.00078 ≈ 0.0008
Pvalue = 0.0008
Since p value, 0.0008 is less than significance level 0.05, we reject H0.
In conclusion, there is sufficient evidence to evidence to reject H0.
b) The type I error here is the claim that less than 50% of adults have a will, whereas the probability is not less than 50%.
Probability of type I error = significance level = 0.05
c)What would your decision be in part A if the probability of making a Type I error were zero?
If probability of making a type I error were zero in part A, we are to reject the null hypothesis, H0.