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

Question

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

Answer 1

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

Answer 2

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.

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

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