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 selected 415 accounting firms and through interviews determines that 303 of these firms ahve flexible scheduling. with a 1% level of significance , does the test show enough evidence to conclude that a significantly lower proportin of accounting firms offer employees flexible scheduling?
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(Hint: this would be a binomial proportion one-sample z-test.)
To determine if the test shows enough evidence to conclude that a significantly lower proportion of accounting firms offer flexible scheduling, we can conduct a hypothesis test.
Step 1: State the hypotheses.
- Null hypothesis (H0): The proportion of accounting firms offering flexible scheduling is the same as or higher than the overall proportion (79%).
- Alternative hypothesis (Ha): The proportion of accounting firms offering flexible scheduling is lower than the overall proportion (79%).
Step 2: Formulate an analysis plan.
- Determine the significance level (alpha): In this case, it is given in the question as 1% or 0.01.
- Conduct the hypothesis test using a one-sample proportion test.
Step 3: Analyze sample data.
In this case, we have the following information:
- Sample size (n): 415 accounting firms
- Number of accounting firms with flexible scheduling (x): 303
Step 4: Calculate the test statistic.
- The test statistic for a one-sample proportion test is calculated using the formula:
z = (p - P) / √(P(1-P) / n)
where:
p is the sample proportion (x/n)
P is the hypothesized proportion (79% or 0.79)
n is the sample size
Plugging in the values:
p = 303/415 = 0.7313
P = 0.79
n = 415
Substituting these values into the formula, we have:
z = (0.7313 - 0.79) / √(0.79 * (1 - 0.79) / 415)
= -3.021
Step 5: Determine the p-value.
- The p-value is the probability of observing a test statistic as extreme as the one calculated under the null hypothesis.
- Since the alternative hypothesis is that the proportion is lower, we are looking for a left-tailed test.
- Using a standard normal distribution table or a calculator, we find that the p-value for z = -3.021 is approximately 0.001.
Step 6: Make a decision.
- Compare the p-value to the significance level (alpha) to make a decision.
- If the p-value is less than alpha (0.01), we reject the null hypothesis.
- In this case, the p-value (0.001) is less than alpha (0.01), so we reject the null hypothesis.
Step 7: State the conclusion.
Based on the results of the hypothesis test, there is enough evidence to conclude that a significantly lower proportion of accounting firms offer employees flexible scheduling compared to the overall proportion.