Question

In preparing a defense against a hostile takeover, a firm’s management assesses the owners of both common and preferred stock with respect to their attitudes toward current management. Of 600 common stockholders, 322 reported being in favor of a change in management. A sample of 220 preferred stockholders found 141 in favor of a management change. The firm’s management is interested in finding evidence at the .05 significance level of a difference in the proportions of common stock holders and preferred stock holders that favor a management change.

1,Which of the following pairs of hypotheses reflect the question of interest?

2.What's the value of the test statistic?

3.What's the p-value for the test statistic in #10?

4.What is the correct decision?

Answer 1

(1)

H0: Null Hypothesis: P1 = P2

HA: Alternative Hypothesis: P1 P2

(2)

So,

Q = 1- P = 0.4354

So,

p1 = 322/600=0.5367

p2 = 141/220 = 0.6409

test statistic is:
Z = (0.5367 - 0.6409)/0.0389 = - 2.68

= 0.05

From Table, critical values of Z = 1.96

Since calculated value of Z = - 2.68 is less than critical value of Z = - 1.96, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim that there is sufficient evidence for significant difference in the proportions of common stock holders and preferred stock holders that favor a management change.

So,

Value of the test statistic is

Z = - 2.68

(3)

Z score = - 2.68

Table of Ara Under Standard Normal Curve gives area = 0.4963

So,

P- value = (0.5 - 0.4963) X 2 = 0.0074

(4)

Since P - Value is less than = 0.05,the difference is significant. Reject null hypothesis.

Correct decision:
There is sufficient evidence for significant difference in the proportions of common stock holders and preferred stock holders that favor a management change.