Excel to Conduct a ttest for Two Samples  Statistics HW Help

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 pvalue 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 Fstatistics, 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:
FTest TwoSample 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) onetail 
0.00834 

F Critical onetail 
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 tstatistics:
\[t=\frac{({{{\bar{X}}}_{1}}{{{\bar{X}}}_{2}})({{\mu }_{1}}{{\mu }_{2}})}{\sqrt{\frac{s_{1}^{2}}{{{n}_{1}}}+\frac{s_{2}^{2}}{{{n}_{2}}}}}\]We use Excel to get the following output:
tTest: TwoSample 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) onetail 
0.012211224 

t Critical onetail 
1.670218808 

P(T<=t) twotail 
0.024422449 

t Critical twotail 
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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