Question

2) Someone you know claims that the average Uber driver makes $4/hour. Let Y be a random variable equal to the hourly earnings of an Uber driver in the U.S. You wish to gather evidence (collect a sample of data) and test the null hypothesis that the average hourly earnings of all Uber drivers in the United States (Hy) is equal to $4 against the alternative hypothesis that it is not equal to $4. Suppose you collect a sample of 400 U.S. Uber drivers, and find that the average earnings of the drivers in your sample is equal to $7 while the variance of earnings in your sample is $900. Finally, suppose that the maximum probability of a Type 1 error that you are willing to tolerate is 5%. (a) State the null and alternative hypotheses (1 points) (b) Calculate that t-statistic (2 points) c) Calculate the p-value (2 points) (d) Compute the 95% confidence interval for My (2 points) (e) Based on either the p-value or the t-statistic, do you reject or fail to reject the null hypothesis? For full credit explain how you arrived to this conclusion. You cannot simply state “Reject" or "Fail to Reject." You need to explain why you are taking the specific action. (2 points)

Answer 1

a) As we are testing here whether the mean is equal to 4, therefore the null and the alternative hypothesis here are given as:

$H_0: \mu = 4$

$H_1: \mu \neq 4$

b) The test statistic here is computed as:

Х – Но 7 – 4 * 2 п 900 400

Therefore 2 is the test statistic value here.

c) For n - 1 = 399 degrees of freedom as this is a two tailed test, the p-value here is obtained from t distribution tables as:
p = 2P( t₃₉₉ > 2) = 2*0.0231 = 0.0462

Therefore 0.0462 is the required p-value here.

d) From standard normal tables, we have here:
P(-1.96 < Z < 1.96) = 0.95

Therefore the confidence interval here is obtained as:

$\bar X \pm z^*\frac{s}{\sqrt{n}}$

$7 \pm 1.96*\sqrt{\frac{900}{400}}$

$7 \pm 2.94$

This is the required 95% confidence interval here.

e) As the p-value here is 0.0462 < 0.05 which is the level of significance, therefore the test is significant here and we Reject the null hypothesis here.