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:
[tex]\bar{x}_A=\frac{\sum x}{N_1}=\frac{12}{6}=2[/tex][tex]\bar{x}_B=\frac{\sum x}{N_2}=\frac{20}{6}=3.33[/tex]
Variance:
[tex]s_A^2=\frac{\sum x^2}{N_1}-\bar{x}_A^2=\frac{28}{6}-2^2=0.667[/tex][tex]s_B^2=\frac{\sum x^2}{N_2}-\bar{x}_B^2=\frac{80}{6}-3.33^2=2.244[/tex]
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.
[tex]H_0:\;\mu_A=\mu_B[/tex]
Research hypothesis:
The research hypothesis is that the mean number of candies handed out by neighborhood A is more than neighborhood B.
[tex]H_a:\;\mu_A>\mu_B[/tex]
Test statistic (t):
The test statistic of a two-sample t-test is given by
[tex]t=\frac{\bar{x}_A-\bar{x}_B}{s_p}[/tex]
Where sp is the pooled standard deviation given by
[tex]\begin{gathered} s_p=\sqrt{\frac{N_1s_1^2+N_2s_2^2}{N_1+N_2-2}(\frac{N_1+N_2}{N_1\cdot N_2}}) \\ s_p=\sqrt{\frac{6\cdot0.667+6\cdot2.244}{6+6-2}(\frac{6+6}{6\cdot6})} \\ s_p=0.763 \end{gathered}[/tex][tex]t=\frac{2-3.33}{0.763}=-1.74[/tex]
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.
