Question

An opinion poll asks a simple random sample of 200 United States immigrants how they view their job prospects. In all, 121 say "good." Does the poll give convincing evidence to conclude that more than half of all immigrants think their job prospects are good? If P "the proportion of all immigrants who say their job prospects are good, what are the hypotheses for a test to answer this question

Answer 1

Solution:

Given: an opinion poll asks a simple random sample of 200 United States immigrants how they view their job prospects.

Thus sample size = n = 200

x = Number of immigrants who say their job prospects are good = 121

We have to test if the poll give convincing evidence to conclude that more than half of all immigrants think their job prospects are good.

We have P = The proportion of all immigrants who say their job prospects are good

Thus we use following steps:

Step 1) State H0 and H1:

H0: P = 0.5    Vs H1: P > 0.5

Step 2) Find test statistic value:

We use z test statistic value for one proportion test.

where

and P = 0.5

Step 3) Find z critical value.

Since level of significance is not given, we use level of significance =

Since H1: P > 0.5 is > type , this is right tailed test.

Thus we use Area = 1 -

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Z_critical = 1.645

Step 4) Decision Rule:

Reject H0, if z test statistic value > z critical value = 1.645, otherwise we fail to reject H0.

Since z test statistic value = 2.97 > z critical value = 1.645, we reject H0.

Step 5) Conclusion:

Since we have rejected H0, the poll gives convincing evidence to conclude that more than half of all immigrants think their job prospects are good.