What is Cheung Chau famous for?
1 answer:
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Answer:
a) The null and alternative hypotheses:
H0 : u ≤ 14
H1 : u > 14
This is a right tailed test.
Let's find test statistics, z
P-value:
Using the standard normal table, NORMSDIST(3.0641) = 0.99889
P(Z>3.0641) = 1 - 0.99889
P(Z>3.0641) = 0.00111
Pvalue = 0.00111
At a significance level of 0.05.
Since pvalue, 0.00111 is less than significance level, 0.05, we reject null hypothesis, H0.
b) Test the hypothesis of part a using the critical-value approach at α = .05
At α = 0.05, (1 - 0.05) = 0.95
Using the standard normal table, the critical value at 0.95 is 1.65.
Reject null hypothesis H0, since Zstat 3.0641 is greater than Zcritical, 1.65.