Question

Dr. Song has spent all her life studying education outcomes and policy in the US. Recently, an article claimed that 6.6% of U. S. students take advanced placement exams. Dr. Song’s work with a random sample of 1,200 students shows that this number is actually 7.5%. Set up Dr. Song’s hypotheses and provide the appropriate test. What do you conclude?

Answer 1

Solution

The solution is based on the Z-test for specific value of the population

Back-up Theory

Normal approximation

X ~ B(n, p) np ≥ 5 and np(1 - p) ≥ 5, (X – np)/√{np(1 - p)} ~ N(0, 1) [approximately]

Now to work out the solution,

Let X = Number of U. S. students take advanced placement exams.

Then, X ~ B(n, p), where n = sample size and p = probability that a U. S. student takes advanced placement exams, which is also equal to the population proportion.

Let p and p_cap respectively be the respective population and sample proportions.

Claim : Proportion of U. S. students taking advanced placement exams is not 0.066 (i.e., 6.6%) as an article claimed.

Hypotheses:

Null H₀ : p = p₀ = 0.066 Vs H_A : p ≠ 0.066

Test Statistic:

Z = (pcap - p₀)/√{p₀(1 - p₀)/n} where pcap = sample proportion and n = sample size.

Calculations:

p0 =	0.066
n =	1200
x =	90
pcap =	0.075
Zcal =	1.2557
α =	0.05 assumed
Zcrit =	1.9600
p-value =	0.2092

Distribution, Significance Level, α Critical Value and p-value:

Under H₀, distribution of Z can be approximated by Standard Normal Distribution, provided

np₀ and np₀(1 - p₀) are both greater than 10.

So, given a level of significance of α%, Critical Value = upper (α/2)% of N(0, 1), and

p-value = P(Z > | Zcal |)

Using Excel Function: Statistical NORMINV and NORMDIST these are found as shown in the above table.

Decision:

Since | Zcal | > Zcrit, or equivalently, since p-value > α, H₀ is accepted.

Conclusion :

There is NOT enough evidence to suggest that the Dr. Song’s contention is valid. Answer

DONE