I think it might be 29.6%
Answer:
b. Based on the answer you found in the above step, provide an argument pro or con to the statement “As the number of women exceed men in the number of bachelor degrees received, expect there to be a corresponding change in the median annual salaries for each gender.”
Step-by-step explanation:
A number of figures are commonly used to describe the gender wage gap. One often-cited statistic comes from the Census Bureau, which looks at annual pay of full-time workers. By that measure, women are paid 80 cents for every dollar men are paid. Another measure looks at hourly pay and does not exclude part-time workers. It finds that, relative to men, typical women are paid 83 cents on the dollar. Other, less-cited measures show different gaps because they examine the gap at different parts of the wage distribution, or for different demographic subgroups, or are adjusted for factors such as education level and occupation.
The presence of alternative ways to measure the gap can create a misconception that data on the gender wage gap are unreliable. However, the data on the gender wage gap are remarkably clear and (unfortunately) consistent about the scale of the gap. In simple terms, no matter how you measure it, there is a gap. And, different gaps answer different questions. By discussing the data and the rationale behind these seemingly contradictory measures of the wage gap, we hope to improve the discourse around the gender wage gap.
840,000. Also, you shouldn't be relying on this site - there are a lot of posts from you in the last couple of minutes. Find out why that's the answer so you can do it yourself!
Answer:
1 1/2
Step-by-step explanation:
60/40 = (60 ÷ 20)/(40 ÷ 20) = 3/2; 3 > 2 => improper fraction Rewrite: 3 ÷ 2 = 1 and remainder = 1 => 3/2 = (1 × 2 + 1)/2 = 1 + 1/2 = = 1 1/2, mixed number (mixed fraction)
Hope this helped!
Answer:
Since the calculated value of z= 1.6279 does not fall in the critical region Z ≥ ±1.96 we conclude that there is no difference between the population variance and the sample variance.
Step-by-step explanation:
The data given is
Population mean μ= $ 7.18
Population variance= σ²= 3.81
Population Standard Deviation = √σ²= √3.81= 1.952
Sample Mean= x`= $ 8.02
Sample Standard Deviation =s = $ 2.08
Sample Size = 15
Significance Level = ∝= 0.05
The null and alternate hypotheses are
H0: σ1=σ2 against the claim that Ha: σ1≠ σ2
where σ1 is the population variance and
σ2 is the sample variance
The rejection region is Z ≥ ±1.96 for two tailed test at ∝= 0.05
The test statistic z is used
z= x`- μ/ σ/√n
Putting the values
Z= 8.02-7.18/1.952/√15
z= 0.84/0.51599
z= 1.6279
Since the calculated value of z= 1.6279 does not fall in the critical region Z ≥ ±1.96 we conclude that there is no difference between the population variance and the sample variance.
