Excel to Conduct a t-test for Two Samples - Statistics HW Help
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Over the last five years statistics teachers have notice a difference in grades for day and night class students. In an attempt to analyze what the differences are and determine the cause two random samples have been taken. Your tasks are to determine if the variance of the populations are equal by running an F Test and then using the appropriate T test given the information from the F test determine, if the night class has a lower mean than the day class.
Using Excel Data Analysis and Word, write a case study. (Must be typed, single space, include data output from excel) After running your excel data you must interpret the analysis.
You report should include an introduction stating what you are trying to determine and why. You should then run descriptive statistics and state what you expect to be the outcome of your hypotheses. Then state your hypotheses using statistical symbols and discuss the data from excel with appropriate critical value rejection rule and p values rejection rules conclude for both p-value and critical value what you find in each situation (Not just means are not equal). The report should be in narrative paragraph form with use of headings and conclusions about what the data tells you.
Solution:
Our purpose is to study the difference observed in grades for day and night class students. The ultimate goal is to compare the means of both populations (day and night students). Since the population variances are unknown, we need first to test for equality of the variances. The outcome of such test will determine the statistics we will use to compare the means.
We select a sample from both populations:
Day |
Night |
36 |
16 |
58 |
30 |
64 |
34 |
71 |
37 |
71 |
40 |
74 |
45 |
75 |
56 |
76 |
61 |
76 |
62 |
77 |
68 |
77 |
69 |
78 |
74 |
79 |
79 |
82 |
80 |
85 |
80 |
86 |
80 |
87 |
80 |
87 |
83 |
88 |
84 |
88 |
84 |
93 |
85 |
93 |
87 |
93 |
87 |
94 |
88 |
94 |
88 |
95 |
88 |
95 |
88 |
96 |
90 |
97 |
91 |
97 |
91 |
98 |
91 |
99 |
92 |
100 |
92 |
100 |
94 |
105 |
98 |
100 |
The descriptive statistics are:
Day |
Night |
||
Mean |
84.68571429 |
Mean |
74.77777778 |
Standard Error |
2.385097792 |
Standard Error |
3.569960898 |
Median |
87 |
Median |
83.5 |
Mode |
93 |
Mode |
80 |
Standard Deviation |
14.11042883 |
Standard Deviation |
21.41976539 |
Sample Variance |
199.1042017 |
Sample Variance |
458.8063492 |
Kurtosis |
2.81565835 |
Kurtosis |
0.699234634 |
Skewness |
-1.349649357 |
Skewness |
-1.280291357 |
Range |
69 |
Range |
84 |
Minimum |
36 |
Minimum |
16 |
Maximum |
105 |
Maximum |
100 |
Sum |
2964 |
Sum |
2692 |
Count |
35 |
Count |
36 |
This information suggests that the mean from day students is higher than the mean from night students.
Variance Comparison:
We have the following hypotheses
\[\begin{array}{cc} & {{H}_{0}}:\sigma _{1}^{2}=\sigma _{2}^{2} \\ & {{H}_{A}}:\sigma _{1}^{2} \not = \sigma _{2}^{2} \\ \end{array}\]We use the F-statistics, defined as
\[F = \frac{s_{1}^{2}}{s_{2}^{2}}\]We perform the analysis with the aid of Excel. The following corresponds to the output:
F-Test Two-Sample for Variances | ||
Night |
Day | |
Mean |
74.77778 |
84.68571 |
Variance |
458.8063 |
199.1042 |
Observations |
36 |
35 |
df |
35 |
34 |
F |
2.304353 |
|
P(F<=f) one-tail |
0.00834 |
|
F Critical one-tail |
2.249408 |
We observe that the significance is equal to \(p = 0.00834\), which is less than 0.01. That means that the test is statistically significant, and we have enough evidence to reject the null hypothesis. Therefore, we can assume that the variances are not equal, at the 0.01 significance level.
Mean Comparison:
Now we have the following hypotheses
\[\begin{array}{cc} & {{H}_{0}}:{{\mu }_{1}}\ge {{\mu }_{2}} \\ & {{H}_{A}}:{{\mu }_{1}}<{{\mu }_{2}} \\ \end{array}\]Since the variances are assumed to be unequal, we use the following t-statistics:
\[t=\frac{({{{\bar{X}}}_{1}}-{{{\bar{X}}}_{2}})-({{\mu }_{1}}-{{\mu }_{2}})}{\sqrt{\frac{s_{1}^{2}}{{{n}_{1}}}+\frac{s_{2}^{2}}{{{n}_{2}}}}}\]We use Excel to get the following output:
t-Test: Two-Sample Assuming Unequal Variances | ||
Night |
Day | |
Mean |
74.77777778 |
84.68571429 |
Variance |
458.8063492 |
199.1042017 |
Observations |
36 |
35 |
Hypothesized Mean Difference |
0 |
|
df |
61 |
|
t Stat |
-2.307711554 |
|
P(T<=t) one-tail |
0.012211224 |
|
t Critical one-tail |
1.670218808 |
|
P(T<=t) two-tail |
0.024422449 |
|
t Critical two-tail |
1.999624146 |
We see from the table that the significance is equal to \(p=0.0122\), which is less than 0.05. That implies that we have enough evidence to reject the null hypothesis, or equivalently, we accept that the grades for day are higher than for night class students, at the 0.05 significance level.
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