A personnel manager claims that 80% of all single women hired for secretarial job get married and
quit work within two years after they are hired. Test this hypothesis at 5% level of significant if among
200 such secretaries, 112 got married within two years after they were hired and quit their job.
To test this hypothesis, we can use a one-sample proportion test.
Null hypothesis (H0): The true proportion of single women hired for secretarial jobs who get married and quit work within two years is 80%.
Alternative hypothesis (H1): The true proportion is not 80%.
Given:
- Number of trials (n) = 200.
- Number of successes (x) = 112.
Calculating the test statistic:
The formula for the test statistic is:
z = (x - np) / sqrt(np(1-p))
First, calculate the expected number of successes under the null hypothesis:
np = 200 * 0.8 = 160.
Next, calculate the expected number of failures under the null hypothesis:
n(1 - p) = 200 * (1 - 0.8) = 200 * 0.2 = 40.
Now, calculate the test statistic:
z = (112 - 160) / sqrt(160 * 40)
z = -48 / sqrt(6400)
z = -48 / 80
z = -0.6
Determining the critical value:
At a 5% level of significance, the critical value for a two-tailed test is approximately ±1.96.
Interpreting the results:
Since the calculated test statistic (-0.6) does not exceed the critical value (-1.96), we fail to reject the null hypothesis. This means that we do not have sufficient evidence to claim that the true proportion of single women hired for secretarial jobs who get married and quit work within two years is not 80%.
Therefore, based on the given data, the personnel manager's claim is not contradicted.