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Answer:
(A) For this research we made use of the paired t test.
(B) For the t test, we conclude that in the null Hypothesis H₀, there was no difference between the average labor force participation rate for women between the years of 1968 and 1972 as against the alternate Hypothesis Hₐ.
From the output data we have, since the P value is 0.024 and is lesser than the alpha value of 0.05, we will accept the alternate Hypothesis instead of the null Hypothesis.
Note:Kindly find an attached copy of the complete question below.
Step-by-step explanation:
Solution
Recall that:
The U.S of Bureau of Labor Statistics have published Values for the labor force participation rate of women (LFPR).
We are more interested in if there was a difference between female participation 1968 and 1972, a rapid change in women.
so we will check values for 19 randomly selected cities for 1968 and 1972,
Now,
(a) Which of this test is most appropriate for this data
The test which will be more suitable for these data will be paired to test t.
Because the data is for labor force participation for two different years, we apply the t test. now that we are making a comparison of two different year times, we will apply the t test in this research.
In this case, the sample size for both the years evaluated should be same, since we need to pair each and all the data.
(b)By using the test you selected, explain your conclusion
Now, to test the data paired t test, we have already the null Hypothesis H₀: In this case there is no difference between the average labor force participation rate for women between the years of 1968 and 1972 as compared with the alternate Hypothesis Hₐ: there was a significant difference between the average labor force participation rate for women between the years of 1968 and 1972.
However, from the output, we have that the P value is 0.024 which is lower than the alpha value of 0.05.
In conclusion, the alternate Hypothesis accepted while the null Hypothesis is rejected.
Note: kindly find an attached copy of the complete question to this solution
below.
Solution :
Given :
Sample mean,
Sample size, n = 129
Sample standard deviation, s = 8.2
a. Since the population standard deviation is unknown, therefore, we use the t-distribution.
b. Now for 95% confidence level,
α = 0.05, α/2 = 0.025
From the t tables, T.INV.2T(α, degree of freedom), we find the t value as
t =T.INV.2T(0.05, 128) = 2.34
Taking the positive value of t, we get
Confidence interval is ,
(32.52, 35.8)
95% confidence interval is (32.52, 35.8)
So with confidence of the population of the mean number of the pounds per person per week is between 32.52 pounds and 35.8 pounds.
c. About of confidence intervals which contains the true population of mean number of the pounds of the trash that is generated per person per week and about
that doe not contain the true population of mean number of the pounds of trashes generated by per person per week.