Question

In the Fall 2019 random sample of 1100 students reviled that 698 of them hold a part time or a full time iob. Let p-population proportion of all students that hold a part time or a full time job this Fall.1) Do we have evidence that proportion of students that hold a part time or a full time iob this Fall is different than 60% which was the propotion last Fall? Use a=0.01
a) Formulate null and alternative hypotheses, use proper symbolic notation.

b) Compute the appropriate test statistics.

c)Give the rejection region for your test, include appropriate sketch, marking critical value(s), and clearly label rejection and non-rejection areas.

d) Compute a p-value for your test, include appropriate sketch.

Answer 1

Solution:

n = 1100

x = 698

Let   be the sample proportion.

= x/n = 0.6345

Let p be the population proportion.

Claim to be tested is " p is different from 0.60

Use = 0.01

/2 = 0.005

a) Null and alternative hypothesis are

H0 : p = 0.60

H1 : p   0.60

b)The test statistic z is

z =   

= (0.6345 - 0.6 0)/[(0.60*(1 - 0.60)/1100]

= 2.34

The appropriate test statistic z is 2.34

c) Now , observe that ,there is   sign in H1. So , the test is two tailed.

So there are two critical values.   i.e.  

i.e.   2.576 (Use z table to find this value)

Critical values are -2.576 and 2.576

Rejection regions : z < -2.576 and z > 2.576

d) Compute a p-value for your test.

For two tailed test :

p value = P(Z < -z) + P(Z > z)

= P(Z < -2.34) + P(Z > 2.34)

= 0.0096 + 0.0096

= 0.0192