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 =
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.
reject the null hypothesis
Oh, I see what you're trying to do here! You're looking for some evidence to support the researcher's belief that accounting firms have a lower proportion of companies offering flexible scheduling. Let me crunch some numbers for you.
To determine if the test shows enough evidence, we can use the formula for the test statistic z:
z = (p̂ - p) / sqrt(p * (1-p) / n)
In this case, p̂ is the sample proportion (303/415), p is the population proportion (79%), and n is the sample size (415).
Plugging in the numbers, we get:
z = ((303/415) - 0.79) / sqrt(0.79 * (1-0.79) / 415)
Calculating this, we find that the test statistic is z = -4.07.
Now, the next step is to compare this test statistic to the critical value from the standard normal distribution at a 1% level of significance. Since you wanted the answer rounded to 2 decimal places, I'll do just that. The critical value for a one-tailed test at a 1% level of significance is -2.33.
Since the test statistic (-4.07) is more extreme than the critical value (-2.33), we can conclude that there is enough evidence to support the claim that a significantly lower proportion of accounting firms offer flexible scheduling.
An accountant with a flexible schedule? Now that's some creative accounting!
The test statistic for testing the difference in proportions is the z-test.
To find the value of the test statistic, we need to compute the z-score using the formula:
z = (p₁ - p₂) / sqrt(p*(1-p)*(1/n₁ + 1/n₂))
Where:
p₁ is the sample proportion (303/415) = 0.7313
p₂ is the population proportion (0.79)
p is the combined proportion [(n₁*p₁ + n₂*p₂) / (n₁ + n₂)]
n₁ is the sample size (415)
n₂ is the population size minus the sample size (N - n₁)
We know that the sample size is 415, but we don't know the population size (N). Since the sample size is relatively small compared to the total number of accounting firms, we can assume N is infinite.
So, p = (n₁*p₁) / (n₁ + n₂) = (415*0.7313) / (415 + ∞) = 0.7313
Now, let's calculate the test statistic using the above formula:
z = (0.7313 - 0.79) / sqrt(0.7313*(1-0.7313)*(1/415 + 1/∞))
z = (0.7313 - 0.79) / sqrt(0.7313*0.2687*(1/415 + 0))
z = -1.9017
Rounded to 2 decimal places, the value of the test statistic is z = -1.90.
To determine whether there is enough evidence to conclude that a significantly lower proportion of accounting firms offer employees flexible scheduling, we can conduct a hypothesis test using the given data.
Let's set up the null and alternative hypotheses:
Null Hypothesis (H0): The proportion of accounting firms offering flexible scheduling is equal to or greater than 79%.
Alternative Hypothesis (H1): The proportion of accounting firms offering flexible scheduling is significantly lower than 79%.
To conduct this hypothesis test, we can use the z-test for proportions.
First, we need to calculate the sample proportion of accounting firms offering flexible scheduling:
Sample Proportion (p̂) = Number of firms with flexible scheduling / Total number of firms = 303/415 = 0.7313
Next, we calculate the standard error of the proportion:
Standard Error (SE) = √[p̂(1 - p̂) / n]
where n is the sample size.
SE = √[(0.7313 * (1 - 0.7313)) / 415] = 0.0212 (rounded to 4 decimal places)
To calculate the test statistic (z), we can use the formula:
z = (p̂ - p) / SE
where p is the proportion stated in the null hypothesis (79%) and p̂ is the sample proportion.
z = (0.7313 - 0.79) / 0.0212 = -2.7708 (rounded to 4 decimal places)
Now, we need to find the critical z-value at the 1% level of significance (α = 0.01) to compare with the test statistic. For a one-sided test, we want to find the z-value that leaves 1% of the area in the right tail.
Using a z-table or a calculator, we find that the critical z-value is approximately -2.33 (rounded to 2 decimal places).
Since the test statistic (z = -2.7708) is more extreme (further into the left tail) than the critical z-value (-2.33), we can reject the null hypothesis.
Therefore, we have enough evidence to conclude that a significantly lower proportion of accounting firms offer employees flexible scheduling.
The value of the test statistic (z) is -2.77.