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 $128 or more over the long run.
Officials initially set rates on the low end,
hoping to attract more shoppers. After a trial
period of 25 days, they will test the claim that
too little revenue is being collected. The
significance level is set at  = 0.05.
Test of a Mean
H0:  = 128
H1:  < 128
Assuming sampling is done randomly:
For a sample from a confirmed Normal population,
or a sufficiently large sample, the test statistic t
follows a(n approximate) t distribution with (n – 1)
DF.
X  X 
t

SX
S n
P-value computations depend on the orientation
(right-, left- two-tailed) of H1.
Data
115.34
123.77
125.44
130.73
99.63
142.43
118.83
110.43
133.67
121.20
137.10
117.48
133.78
119.11
117.66
123.65
141.30
96.86
137.56
121.98
X  122.50 S  13.50
87.00
128.54
121.94
124.70
132.37
Example
7
Frequency
6
5
4
3
2
1
0
80
96
112
128
144
Daily Parking Revenue ($)
X  122.50
 S  13.50
H0:  = 128
Test Statistic / P-value
H0:  = 128
H1:  < 128
X  122.50 S  13.50 n  25
X   122.5  128.0  5.50
t


 2.037
2.70
S n
13.5 25
Since the test is left tailed, we need the area
below -2.037 for a t distribution with 24 DF.
Sampling Distribution of T24 When
H0 is True
From Data
From Table
Sampling Distribution of T24 When
H0is True
Just a bit more than 0.025
From Data
From Table
Test Statistic / P-value
X   122.5  128  5.5
t


 2.037
2.61
S n 13.5 25
Since the test is left tailed, we need the area below
-2.037 for a t distribution with 24 DF.
That area is between 0.025 and 0.05 and much
closer to 0.025.
Using Minitab we get: P-value = 0.026.
Corraboration
T = -2.037
The critical value for a 5% test is –t0.05 = -1.711.
A 90% CI for :
E = 1.711 2.61 = 4.47
118.0 <  < 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 -2.037
(which comes from the sample mean of $122.50,
with standard deviation $13.50).
Conclusion
Since P-value = 0.026 < 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
H0:  = 30
H1:  > 30
n = 19
DF = 18
T = 1.96
P-value = 0.033
True or false?
3.3% of the data is above 30.
FALSE
Quiz
H0:  = 30
H1:  > 30
n = 19
DF = 18
T = 1.96
P-value = 0.033
True or false?
3.3% of the data is below 30.
FALSE
Quiz
H0:  = 30
H1:  > 30
n = 19
DF = 18
T = 1.96
P-value = 0.033
True or false?
3.3% of the cars go 30.
FALSE
Quiz
H0:  = 30
n = 19
H1:  > 30
DF = 18
T = 1.96
P-value = 0.033
True or false?
The probability the null hypothesis is true is
0.033.
FALSE
but if you think this way you’ll make
correct decisions
What a P-value is…
H0:  = 30
n = 19
T = 1.96
H1:  > 30
DF = 18
P-value = 0.033
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
0.033.
T = 1.96 because the sample mean is 1.96 standard
deviations above 30. Not so likely.