Null hypothesis:
Ho: p = .79 -->meaning: population proportion is equal to .79
Alternative hypothesis:
Ha: p < .79 -->meaning: population proportion is less than .79
Using a formula for a binomial proportion one-sample z-test with your data included, we have:
z = (.73 - .79) -->test value (303/415 = .73) minus population value (.79)
divided by
√[(.79)(.21)/415] --> .21 represents 1 - .79 and 415 is the sample size.
Use a z-table to find the critical or cutoff value for a one-tailed test (lower tail) at .01 level of significance. The test is one-tailed because the alternative hypothesis is showing a specific direction (less than).
If the test statistic exceeds the critical value you find from the table, reject the null. If the test statistic does not exceed the critical value from the table, do not reject the null.
You can draw your conclusions from there.
I hope this will help get you started.
A study by Hewitt Associates showed that 79% of companies offer employees flexible scheduling. Suppose a researcher believes that in accounting firms this figure is lower. The researcher randomly selects 415 accounting firms and through interviews determines that 303 of these firms have flexible scheduling. With a 1% level of significance, does the test show enough evidence to conclude that a significantly lower proportion of accounting firms offer employees flexible scheduling?
Round your answer to 2 decimal places, the tolerance is +/-0.05.
The value of the test statistic is z =
2 answers
reject the null hypothesis
1 answer