The objective of this is to carry out a hypothesis test using the Welch Approximate t Procedure, to determine if there is evidence of a difference in the caffeine content between sugar and diet soda.
** I know longer type question so posted it 3-4 times to help :) **
A) Use the Data Analysis ToolPak to carry out the Welch Approximate t Procedure. Include the output from Excel as a figure :
B) Identify the value of the test statistic from the Excel output :
C) Identify the p-value using the Excel output, and use it to determine whether or not the null hypothesis is rejected, assuming a 5% level of significance :
D) Calculate a 95% confidence interval for M1 - M2. Clearly state the point estimate for M1 - M2, the confidence coefficient, and the final confidence interval. To help with your calculations, use the rounded degrees of freedom that are reported as part of the Excel output. :
E) Provide an appropriate interpretation of the confidence interval you created. Does this interval support the results of your hypothesis test? Explain why or why not.
F) State the assumptions that must be satisfied for the inference procedures you just carried out to be valid. Using the plots you created as well as information provided in the question, is there reason to believe these assumptions have been violated? Justify your response
G) State a final conclusion about the caffeine content in sugar versus diet soda, using the results of this analysis :
Sugar |
Diet |
28.43 |
29.24 |
29.38 |
29.32 |
29.93 |
30.83 |
30.56 |
31.18 |
30.64 |
31.78 |
32.04 |
31.78 |
32.3 |
31.86 |
32.56 |
31.87 |
32.61 |
32.05 |
32.73 |
32.08 |
32.98 |
32.12 |
33.04 |
32.24 |
33.23 |
32.79 |
33.24 |
32.83 |
33.41 |
32.97 |
33.44 |
32.98 |
33.46 |
33.11 |
33.55 |
33.11 |
33.57 |
33.44 |
33.79 |
33.57 |
34 |
33.69 |
34.01 |
33.85 |
34.09 |
34.02 |
34.15 |
34.27 |
34.26 |
34.54 |
34.26 |
34.83 |
34.27 |
34.87 |
34.3 |
35.03 |
34.36 |
35.39 |
34.38 |
35.53 |
34.46 |
35.54 |
34.59 |
35.54 |
34.61 |
35.64 |
34.85 |
35.69 |
34.88 |
36.1 |
35.01 |
36.1 |
35.08 |
36.51 |
35.34 |
36.85 |
35.4 |
36.92 |
35.59 |
37.19 |
35.66 |
37.49 |
35.68 |
37.6 |
35.74 |
38.06 |
35.81 |
38.27 |
35.88 |
38.32 |
35.91 |
38.56 |
36.14 |
38.57 |
36.19 |
39.07 |
36.22 |
39.09 |
36.31 |
39.13 |
36.32 |
39.34 |
36.32 |
39.39 |
36.43 |
39.59 |
36.54 |
39.71 |
36.62 |
39.82 |
36.63 |
39.85 |
36.79 |
39.87 |
36.8 |
40.07 |
36.81 |
40.1 |
37.03 |
40.11 |
37.1 |
40.25 |
37.21 |
41.06 |
37.44 |
41.07 |
37.49 |
41.27 |
37.82 |
41.27 |
37.85 |
41.5 |
37.94 |
42.05 |
37.97 |
42.55 |
38.43 |
43.8 |
38.47 |
43.93 |
38.6 |
44.21 |
38.72 |
44.35 |
38.75 |
44.39 |
38.81 |
44.5 |
38.82 |
44.6 |
38.88 |
44.71 |
38.9 |
44.76 |
39.01 |
45.18 |
39.01 |
45.19 |
39.06 |
45.66 |
39.13 |
45.77 |
39.3 |
45.91 |
39.32 |
46 |
39.5 |
46.21 |
39.66 |
46.37 |
39.8 |
46.55 |
39.93 |
46.65 |
40 |
46.73 |
40.04 |
46.85 |
40.06 |
46.85 |
40.35 |
46.99 |
40.37 |
47.26 |
40.38 |
47.26 |
40.45 |
47.47 |
40.47 |
47.49 |
40.9 |
47.52 |
41.05 |
47.77 |
41.19 |
47.79 |
41.24 |
47.87 |
41.36 |
47.91 |
41.4 |
47.95 |
41.92 |
48.06 |
42.09 |
48.11 |
42.16 |
48.25 |
42.18 |
48.27 |
42.28 |
48.35 |
42.48 |
48.47 |
42.86 |
48.7 |
43.25 |
48.97 |
43.52 |
48.98 |
43.78 |
49.12 |
43.86 |
49.31 |
44 |
49.37 |
45.16 |
49.64 |
45.17 |
49.65 |
45.96 |
49.71 |
46.42 |
50.74 |
47.32 |
50.82 |
47.7 |
51.2 |
48.12 |
52.54 |
A) Using excel, running a Welch Approximate t Procedure:
Let
denote the mean caffeine content in Sugar and Diet soda. We have
to test:
Vs
We get the output:
B) From the output, the test statistic is obtained as -5.008.
C) The p-value of the test is obtained as 0.000. Comparing the wih p-value with the significance level 0.05; Since, p-value = 0.000< 0.05, we may reject H0 at 5% level. We may conclude that there is a significant difference in the caffeine content in Sugar and Diet soda.
(D) The 95% CI for difference in mean
is given by the formula:
where,
Substituting the values,
=( -2.069, -4.730)
Since the 95% CI does not contain the null value (),
we may conclude that the interval is significant. We arrive at the
same conclusion as from the t test: Reject H0. We may conclude that
there is a significant difference in the caffeine content in Sugar
and Diet soda.
(E) Assumptions of Welch t test:
- The data (observations of both the independent groups) is approximately normally distributed - The data collected is random
Checking the assumptions: We may check the normality assumption using a Histogram:
Similarly, for diet:
Since the bell shaped graph is
just a crude measure of normality, we might want to ensure that the
condition is definitely satisfied. For this we may go for the
summary statistics. If the mean and median of the population are
almost equal, we would have a stronger evidence.
We may conclude that the normality assumption is satisfied.
(G) We have run the appropriate test for equality of means and have also ensured that the assumptions of the test are satisfied. Thus, based on the analysis, we may conclude that: There is a significant difference in the caffeine content in Sugar and Diet soda.
The objective of this is to carry out a hypothesis test using the Welch Approximate t...