2x + y = -4 5x - 3y = 1 Solve with elimination method
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We are given the data on the number of candies handed by neighborhood A and neighborhood B.
Let us first find the mean and variance of each neighborhood.
Mean:
Variance:
A. Null hypothesis:
The null hypothesis is that there is no difference in the mean number of candies handed out by neighborhoods A and B.
Research hypothesis:
The research hypothesis is that the mean number of candies handed out by neighborhood A is more than neighborhood B.
Test statistic (t):
The test statistic of a two-sample t-test is given by
Where sp is the pooled standard deviation given by
So, the test statistic is -1.74
Critical t:
Degree of freedom = N1 + N2 - 2 = 6+6-2 = 10
Level of significance = 0.05
The right-tailed critical value for α = 0.05 and df = 10 is found to be 1.81
Critical t = 1.81
We will reject the null hypothesis because the calculated t-value is less than the critical value.
Interpretation:
This means that we do not have enough evidence to conclude that neighborhood A gives out more candies than neighborhood B.