Question

Paired t‐Test for Mean Comparison with Dependent Samples To study the effects of an advertising campaign at a supply chain, several stores are randomly selected with the following observed before‐ and after‐advertising monthly sales revenues:
Store number 1 2 3 4 5
Old sales revenue (mil. $) 5.2 6.5 7.2 5.7 7.6
New sales revenue (mil. $) 6.4 7.8 6.8 6.5 8.2
Let μ₁ and μ₂ be the means of old and new sales revenues, both in millions of dollars per month.
(a) At α = 0.05, test H0: μ2 ≤ μ1 versus H1: μ2 > μ1. Sketch the test. Interpret your result.
(b)Sketch and find the p‐value of the test. Would you reject H0 if α = 0.01?

Answer 1

Answer:

a) $t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.7 -0}{\frac{0.678}{\sqrt{5}}}=2.308$  

$p_v =P(t_{(4)}>2.308) =0.0411$

So the p values is lower than the significance level given 0.05, so then we can conclude that we reject the null hypothesis.

b) The p value is illustrated on the figure attached.

If we select $\alpha=0.01$ we see that $p_v >\alpha$ so then we have enough evidence to FAIL to reject the null hypothesis.

Step-by-step explanation:

Part a

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

1=test value old , 2 = test value new

1: 5.2 6.5 7.2 5.7 7.6

2: 6.4 7.8 6.8 6.5 8.2

The system of hypothesis for this case are:

Null hypothesis: $\mu_2- \mu_1 \leq 0$

Alternative hypothesis: $\mu_2 -\mu_1 >0$

The first step is calculate the difference $d_i=y_i-x_i$ and we obtain this:

d: 1.2, 1.3, -0.4, 0.8, 0.6

The second step is calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}=0.7$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =0.678$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.7 -0}{\frac{0.678}{\sqrt{5}}}=2.308$

The next step is calculate the degrees of freedom given by:

$df=n-1=5-1=4$

Now we can calculate the p value, since we have a right tailed test the p value is given by:

$p_v =P(t_{(4)}>2.308) =0.0411$

So the p values is lower than the significance level given 0.05, so then we can conclude that we reject the null hypothesis.  

Part b

The p value is illustrated on the figure attached.

If we select $\alpha=0.01$ we see that $p_v >\alpha$ so then we have enough evidence to FAIL to reject the null hypothesis.