Question

In: Statistics and Probability

The opinion of university alumni is important. One question from a recent survey of 450 business...

The opinion of university alumni is important. One question from a recent survey of 450 business alumni was, “Should foreign language be required of all undergraduate business majors?” The results of the study indicated that 30% of the sample were in favor of a foreign language requirement; 54% were opposed to a foreign language requirement; and the remainder were undecided. Do the survey results indicate that the majority of alumni do not support a foreign language requirement? Solve manually and then verify your results with PHStat. Include Type II error with your solution.

Solutions

Expert Solution

Solution:

Here, we have to use z test for population proportion. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: Majority of alumni support a foreign language requirement.

Alternative hypothesis: Ha: Majority of alumni do not support a foreign language requirement.

H0: p ≥ 0.5 versus Ha: p < 0.5 (one tailed test)

This is lower tailed or left tailed test.

If proportion of alumni is 0.5 or more support the foreign language requirement, then we can say that there is a majority. Majority means more than half or at least half.

We are given

Sample size = n = 450

Sample proportion = P = 30% = 0.30

We assume α = 0.05

Test statistic is given as below:

Z = (P – p) / sqrt[p*(1 – p)/n]

Z = (0.30 – 0.50) / sqrt(0.5*(1 – 0.5)/450)

Z = -0.20 / 0.0236

Z = -8.4853

Critical value = -1.6449

(by using z-table)

P-value = 0.0000

(by using z-table)

P-value < α = 0.05

So, we reject the null hypothesis that Majority of alumni support a foreign language requirement.

There is sufficient evidence to conclude that Majority of alumni do not support a foreign language requirement.

Type II error = P(Do not reject H0 | H0 is not true) = 0.00

PHStat output is given as below:

Z Test of Hypothesis for the Proportion

Data

Null Hypothesis            p =

0.5

Level of Significance

0.05

Number of Items of Interest

135

Sample Size

450

Intermediate Calculations

Sample Proportion

0.3

Standard Error

0.0236

Z Test Statistic

-8.4853

Lower-Tail Test

Lower Critical Value

-1.6449

p-Value

0.0000

Reject the null hypothesis

From this output, it is observed that all values are correct.


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