Test for a Mean

Example A city needs $32,000 in annual revenue from parking fees. Parking is free on weekends and holidays; there are 250 days in which parking is not free. This implies that the daily revenue must average $128 or more over the long run. Officials initially set rates on the low end, hoping to attract more shoppers. After a trial period of 25 days, they will test the claim that too little revenue is being collected. The significance level is set at  = 0.05.

Test of a Mean H 0 :  = 128H 1 :  < 128 Assuming sampling is done randomly: For a sample from a confirmed Normal population, or a sufficiently large sample, the test statistic t follows a(n approximate) t distribution with (n – 1) DF. P-value computations depend on the orientation (right-, left- two-tailed) of H 1.

Data

Example H 0 :  = 128

Test Statistic / P-value Since the test is left tailed, we need the area below for a t distribution with 24 DF. H 0 :  = 128H 1 :  < 128

Sampling Distribution of T 24 When H 0 is True From Table From Data

Sampling Distribution of T 24 When H 0 is True From Table From Data Just a bit more than 0.025

Test Statistic / P-value Since the test is left tailed, we need the area below for a t distribution with 24 DF. That area is between and 0.05 and much closer to Using Minitab we get: P-value =

Corraboration T = The critical value for a 5% test is –t 0.05 = A 90% CI for  :E =  2.61 = <  < 127.0

Interpretation of P-value Suppose revenue has population mean $128 per day. (Only) 2.6% of all possible samples of size 25 give a T statistic as low as the observed (which comes from the sample mean of $122.50, with standard deviation $13.50).

Conclusion Since P-value = < 0.05 we reject the null hypothesis. Simple, non technical interpretation of the conclusion… There is sufficient evidence in the data to conclude that the mean daily revenue is less than the needed $128 per day.

Quiz – Based on Worksheet H 0 :  = 30H 1 :  > 30 n = 19DF = 18 T = 1.96P-value = True or false? 3.3% of the data is above 30. FALSE

Quiz H 0 :  = 30H 1 :  > 30 n = 19DF = 18 T = 1.96P-value = True or false? 3.3% of the data is below 30. FALSE

Quiz H 0 :  = 30H 1 :  > 30 n = 19DF = 18 T = 1.96P-value = True or false? 3.3% of the cars go 30. FALSE

Quiz H 0 :  = 30H 1 :  > 30 n = 19DF = 18 T = 1.96P-value = True or false? The probability the null hypothesis is true is FALSE but if you think this way you’ll make correct decisions

What a P-value is… H 0 :  = 30H 1 :  > 30 n = 19DF = 18 T = 1.96P-value = Assume (pretend) the null hypothesis is true. Consider (pretend) the study is redone. The probability of a T at least as high as 1.96 is T = 1.96 because the sample mean is 1.96 standard deviations above 30. Not so likely.