Question

Problem #1 -- Historically, 20% of graduates of the engineering school at a major university have been women. In a recent, randomly selected graduating class of 210 students, 58 were females. Does the sample data present convincing evidence that the proportion of female graduates from the engineering school has shifted (changed)? Use α = 0.05.

A. Explain what it means to make a Type I error and indicate the probability of it occurring, under the assumption the null hypothesis was true.

B. Explain what it means to make a Type II error and suggest a possible value of the parameter in order for such an error to occur.

Answer 1

We have to test the hypothesis:

H₀ : p = 0.20

H₁ : p 0.20

We would perform a two sided test.

The test statistic under H₀ is given by:

Z =  

i.e.

Z =

i.e.

Z = 2.76

Now at 5% level of significance for a two tailed test:

Z_0.025 = 1.96

Now since Z = 2.76 > Z_0.025 = 1.96, therefore we reject null hypothesis H_o at 5% level of significance and conclude that sample data provide convincing evidence that the proportion of female graduates from the engineering school has shifted (changed).

A.

Type I error is the rejection of a true null hypothesis i.e. P(Reject H₀ / H_o is true)

Here P(Reject H₀ / H_o is true) = 5% = 0.05

B.

Type II error is failing to reject a false null hypothesis i.e. P(Accept H₀ / H_o is false)

Here P(Accept H₀ / H_o is false) = = 0.211

Conceptual information:

Where

• n is sample size
• p₀ is the comparison value
• Φ is the standard Normal distribution function
• α is Type I error
• β is Type II error, meaning 1−β is power

Used R to solve the same. Code being :

p=0.27619

p0=0.2

alpha=0.05

ceiling(n) # 210

z=(p-p0)/sqrt(p*(1-p)/n)

(Power=pnorm(z-qnorm(1-alpha/2))+pnorm(-z-qnorm(1-alpha/2)))