Question

An online stock trading company makes part of their revenue from clients when the clients trade stocks therefore, it is important to the company to have an good idea of how many trades its clients are making in a given year. In a sample of 103 clients of an online stock trading company, the average number of trades per year was 86 with a standard deviation of 16. If you were to test the hypothesis that theaverage number of trades per year is different than the previous year when the average number of trades was 84 (using the % level of significance), what is the test statistic? (please round your answer to 2 decimal places)

Answer 1

Here we are testing whether the mean number of trades per year is different from 84, therefore the null and the alternate hypothesis here is given as:

$H_0: \mu = 8 4$

$H_a: \mu \neq 8 4$

From the given sample statistics, the test statistic here is computed as:

$t^* = \frac{\bar X - \mu_0 }{\frac{s}{\sqrt{n}}} = \frac{86 - 84}{\frac{16}{\sqrt{103}}} = 1.2686$

Now as this is a two tailed test, the p-value here is computed from the t distribution tables for n-1 = 102 degrees of freedom as:

p = 2P( t₁₀₂ > 1.2686 ) = 2*0.1037 = 0.2074

Here as the p-value is very large and greater than the level of significance for the test, therefore the test is not significant and we cannot reject the null hypothesis here. Therefore we dont have sufficient evidence here that the average number of trades per year is different than the previous year when the average number of trades was 84