Refer to Exhibit 95 If the test is done at a 2 level of sign

Refer to Exhibit 9-5. If the test is done at a 2% level of significance, the null hypothesis should:

A random sample of 100 people was taken. Eighty of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 75%.

Refer to Exhibit 9-6. At a .05 level of significance, it can be concluded that the proportion of the population in favor of candidate A is:

d. significantly greater than 75%.

A random sample of 16 students selected from the student body of a large university had an average age of 25 years. We want to determine if the average age of all the students at the university is significantly different from 24. Assume the distribution of the population of ages is normal with a standard deviation of 2 years.

Refer to Exhibit 9-4. At a .05 level of significance, it can be concluded that the mean age is:

Exhibit 9-5 Assume population is normally distributed.

Solution

1)

Test static = (X-bar - Mu)/(SD/sqrt(n))

Test static = (75.607 - 80)/(8.246/sqrt(16)) = -2.13

P-value (X>80) = 1 - 0.0166 = 0.9834 > 0.02

Hence null hypothesis should not be rejected. Option d

2)

p = 0.8

P=0.75

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion,

Test Static = (p -P)/sqrt(P(1-P)/n)

=(0.8 -0.75)/sqrt(0.75(1-0.75)/100)

=1.15

P-value ( p>0.75) = 1 - 0.8749 = 0.1251>0.05

Hence ,

The population in favor of candidate A is significantly greater than 75%. Option d

3)

Test static = (X-bar - Mu)/(SD/sqrt(n))

Test static = (25 - 24)/(2/sqrt(16)) = 2

P-value (X>80) =2( 1 - 0.9772 )= 2* 0.0228 = 0.0456< 0.05

Mean age is significantly different from 24. Option a