Question

I need to know how to do the whole thing and understand it.

Answer 1

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:

$\bar{x}_A=\frac{\sum x}{N_1}=\frac{12}{6}=2$$\bar{x}_B=\frac{\sum x}{N_2}=\frac{20}{6}=3.33$

Variance:

$s_A^2=\frac{\sum x^2}{N_1}-\bar{x}_A^2=\frac{28}{6}-2^2=0.667$$s_B^2=\frac{\sum x^2}{N_2}-\bar{x}_B^2=\frac{80}{6}-3.33^2=2.244$

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.

$H_0:\;\mu_A=\mu_B$

Research hypothesis:

The research hypothesis is that the mean number of candies handed out by neighborhood A is more than neighborhood B.

$H_a:\;\mu_A>\mu_B$

Test statistic (t):

The test statistic of a two-sample t-test is given by

$t=\frac{\bar{x}_A-\bar{x}_B}{s_p}$

Where sp is the pooled standard deviation given by

$\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}$$t=\frac{2-3.33}{0.763}=-1.74$

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.