# Is my Answer to: Is the Price in NY is 20% higher than that in LA, right?

Is my Answer to: Is the Price in NY is 20% higher than that in LA, right?

H0: μNY -μLA >= 0.20 (20%)

while

H1: μNY - μLA < 0.20 (20%)

Formula:

T score= (Percentage mean difference - hypothesized mean difference )/ standard error

Percentage mean difference = (Sample mean NY - Sample Mean LA) / Sample Mean LA

= (3.94 - 3.25) / 3.25 = 0.69 / 3.25 = 0.212

T score = ((0.69 / 3.25) - 0.20) / 0.11

T score = 0.112

t_statistic for 10 + 8 -2 = 16 degrees of freedom at 0.05 significance, one-sided test = 1.746

T score < t statistic hence accept the null hypothesis

p-value = 0.4561 > 0.05 hence accept the null hypothesis

p-value = 0.4561 > 0.01 hence accept the null hypothesis at 0.01 significance level

Therefore the price of apples in NY is statistically proven to be higher than 20% the price of apples in LA at all common and many uncommon significance levels

I may be wrong, but I think you mixed up H0 and H1. Accepting the null doesn't prove anything, only rejecting does. https://www.youtube.com/watch?v=bf3egy7TQ2Q

I agree with Klim.

Further, the T-score should be bigger than 2, to reject the null hypothesis (rule of thumb).

Calculating with absoulte values ( $ ), I got :

- d = μNY - μLA = $0.7 as sample means and

- μ0 = $3.25 * 20% as hypothesized mean difference

- T-score: T = d-μ0/Std error = 0.7-0.65/0.11 = 0.44

- t_16_0.05 = 1.746

-> accept the null hypothesis -> there is not enoug statistical evidence that the NY apples are 20% more expensive.

I think that it should be two-sided test again- the way he asked the question:

Is price in NY 20% higher than the price in LA or not?

Ho: price is 20% higher and

Ha: price is **not **20% higher,

so we try to prove it is not 20% higher.

The way I did it is just substituted the 8 sample values fro LA with 20% higher values in the excel table, which is attached. You get results

3.624

3.864

3.888

3.624

3.672

3.78

4.572

4.128

and then the formulas are already enterred in the excel table, so you get:

Pooled variance 0.06

Pooled std 0.12

T-score 0.39

I am not sure if thats the correct approach though