Respuesta :
Answer:
Step-by-step explanation:
Hello!
You have two study variables
X₁: Time it takes to get to school by bus.
X₂: Time it takes to get to school by car.
Data:
Sample 1
Bus:(15,10,7,13,14,9,8,12,15,10,13,13,8,10,12,11,14,11,9,12)
n₁= 20
Mean X[bar]₁= 11.30
S₁= 2.39
Sample 2
Car:(5,8,7,6,9,12,11,10,9,6,8,10,13,12,9,11,10,7)
n₂= 18
Mean X[bar]₂= 9.06
S₂= 2.29
A.
To test if there is any difference between the times it takes to get to school using the bus or a car you need to compare the means of each population.
The condition needed to make a test for the difference between means is that both the independent population should have a normal distribution.
The sample sizes are too small to use an approximation with the CLT. You can test if the study variables have a normal distribution using different methods, and hypothesis test, using a QQ-plot or using the Box and Whiskers plot. The graphics are attached.
As you can see both samples show symmetric distribution, the boxes are proportioned, the second quantile (median) and the mean (black square) are similar and in the center of the boxes. The whiskers have the same length and there are no outliers. Both plots show symmetry centered in the mean consistent with a normal distribution. According to the plots you can assume both variables have a normal distribution.
The next step to select the statistic to test the population means is to check whether there is other population information available.
If the population variances are known, you can use the standard normal distribution.
If the population variances are unknown, the distribution to use is a Student's test.
If the unknown population variances are equal, you can use a t-test with a pooled sample variance.
If the unknown population variances are not equal, the t-test to use is the Welch approximation.
Using an F-test for variance homogeneity the p-value is 0.43 so at a 0.01 level, you can conclude that the population variances are equal.
The statistic to use is a pooled t-test.
B.
Degrees of freedom.
For each study variable, you can use a t-test with n-1 degrees of freedom.
For X₁ ⇒ n₁-1 = 20 - 1 = 19
For X₂ ⇒ n₂-1 = 18 = 17
For X₁ + X₂ ⇒ (n₁-1) + (n₂-1)= n₁ + n₂ - 2= 20 + 18 - 2= 36
C.
See above.
D.
The formula for the 99% confidence interval is:
(X[bar]₁ - X[bar]₂) ± [tex]t_{n_1+n_2-2; 1- \alpha /2}[/tex] * [tex]Sa\sqrt{\frac{1}{n_1} + \frac{1}{n_2} }[/tex]
[tex]Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} }[/tex]
[tex]Sa= \sqrt{\frac{19*(2.39)^2+17*(2.29)^2}{36} }[/tex]
Sa= 2.34
[tex]t_{n_1+n_2-2; 1- \alpha /2}[/tex]
[tex]t_{36; 0.995}[/tex] = 2.72
(11.30 - 9.06) ± 2.72 * [tex]2.34\sqrt{\frac{1}{20} + \frac{1}{18} }[/tex]
[0.17;4.31]
With a 99% confidence level you'd expect that the difference between the population means of the time that takes to get to school by bus and car is contained in the interval [0.17;4.31].
E.
Couldn't find the original lesson to see what calculator is used.
F.
Same, no calculator available.
I hope it helps!
Answer:
this is nnot the answer i was looking for
Step-by-step explanation:
