Question

You have a falafel cart and you sell falafel every weekday near Washington Square Park during lunch time. Your daily revenue is normally distributed with a mean of $200 and a standard deviation of $50.

(a) Suppose there is another location that might be worth switching to. You plan to experiment with selling there for awhile, and then use a hypothesis test to determine whether you should switch. If the new location has a normally distributed revenue with a true mean of 210 and a standard deviation of 50, how many days would you have to try selling there to have a power of 50%. Use an α = .05 (significance level).

(b) Suppose you try selling at another location for 16 days, and on average you sell $220 worth of falafel with a sample standard deviation of $36 Using an α = .05, test whether the new location is worth switching to.

Answer 1

(A)

New location will be worth if revenue seems to be increase in new location so we need to test whether average revenue is greater than 200. That is test will be right tailed.

Here we have σ-50, CL-0.95, β-0.5 Level of significance: α-0.05 Power-0.5 Power: Test is one tailed so critical values are: 1-a1.645 δ-14-61-10 Delta: Required sample size is: n-1.645+0)*50)/10)A2 67.650625 Hence, the required sample size is 68. Excel function for critical value: -NORMSINV(1-(0.05/2)) -NORMSINV(1-05) 4-α 71-β

(B)

Here we have 220, s -36, n -16 Hypotheses are H200 Level of significance a-0.05 Test is one tailed (right tailed) df -n-1-15 Degree of freedom: Test Statistics: - (220-200)/(36/sqrt(16)) -2.222 Critical value: Rejection Region 1.753 If |t| >1.753, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value P-value-0.021 Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: Excel function for p-value: TINV(0.1,15) -TDIST(2.222,15,1)