OC3140
HW/Lab 8 Hypothesis Testing
1. Is the sea surface temperature in September in the Gulf of Mexico significantly
hotter than 25o C ? A sample of 10 days of SST was taken (sample size n = 10).
The sample mean and standard deviation are 27o C and 2o C , respectively.
Solution:
This is the Hypothesis Testing on the mean (see Ch. 7 p7-p8) (as small sample
and unknown  )

Start on the null hypothesis: H 0 : sst  25o C
Define the alternative hypothesis: H A : sst  25o C
Identify a test statistic: for a testing on means, a t distribution is used:
x   sst
t
s
n
 Identifying a critical value tc with   0.05, d . f .  n  1  9 and H A is
upper one-side, we have (from the t table Ch.5 p17)
tc  1.833 .
 For the testing sample,
n=10, mean=27, standard deviation = 2,
the t-value is
27  25
t
 3.1623 .
2
10
 Since t   3.1623  tc   1.833 , we reject the null hypothesis H 0 and


conclude that sea surface temperature is significantly hotter than 25o C .
2. A marine science equipment needs dozens uniform batteries. The manufacturer
claims a variance of 0.012. Sample size = 50 batteries, sample variance=0.02.
Determine if the manufacture’s claim can be accepted.
Solution:
This is the Hypothesis Testing on the Variance (see Ch. 7, p8-p9)

Start on the null hypothesis: H 0 :  2  0.012


Define the alternative hypothesis:
Identify the test statistics,
H A :  2  0.012
 
2

 n  1 s 2
2
Identifying a critical value base on:
  0.05, d . f .  n  1  49 and H a :  2  0.012 (i.e., upper one-sided),
we have (from the  2 table Ch.5 p12),
2
c2  0.05,49
 67.5 .


Compute the test statistics from the sample:
 n  1 s 2 49  0.02
2 

 81.67 .
0.012
2
Since  2   81.67   c2   67.5 , we reject H 0 and conclude that the
batteries are significantly more variable than the manufacture claimed.
3. The Edison Electric Institute has published figures on the annual number of
kilowatt-hours expended by various home appliances. It is claimed that a vacuum
cleaner expends an average of 46 kilowatt-hours per year. If a random sample of
12 homes included in a planned study indicates that vacuum cleaners expend an
average of 42 kilowatt-hours per year with a standard deviation of 11.9 kilowatthours, does this suggest at the 0.05 level of significance that vacuum cleaners, on
the average, less than 46 kilowatt-hours annually? Assume the population of
kilowatt-hours to be normal.
Solution:
This is the Hypothesis Testing on the mean (see Ch. 7 p7-p8) (as small sample
and unknown  )
 Start on the null hypothesis: H 0 :   46kilowatt  hours
 Define the alternative hypothesis: H1 :   46kilowatt  hours
 Identify a test statistic: for a testing on means, a t distribution is used:
x  0
t
s
n
 Identifying a critical value base on:
  0.05, d . f .  n 1  11 and the critical t0.05  1.796 , (as x  0 and
t  0 , so lower one-sided will be used).
x  0
42  46
 t

 1.16  t0.05
s
11.9 / 12
n
 Decision: Do not reject H 0 and conclude that the average number of kilowatt-hours expended annually by home vacuum cleaner is not
significantly less than 46.
4. There are some temperature samples of this month in Monterey city and Marina
city. The sample size in Monterey is 25 and the ample size in Marina city is 17.
The sample means and variances were computed and tabulated in the following.
Sample Size
Sample mean
Sample Variance
Monterey
25
75.5
4
Marina
17
73.3
6
Is there a significant difference of the mean temperature between Monterey and
Marina cities at a significance level of 5%?
Solution:
This is the Hypothesis Testing on the difference of two means (see Ch. 7 p10p12)
State the null hypothesis: H o : 1  2
Define the alternative hypothesis: H A : 1  2
Identify the test statistics: t distribution.



Method: 1 assume 1   2
 Identifying a critical value: d . f .  n1  n2  2  40 ,   0.05 and it is
two-sided testing, we have t  2.0211
2
,40

Compute the t value from the sample
x1  x2
75.5  73.3
t

 3.1943 where
1
 24 * 4  16 * 6   1
1
2  1
s   

  
40

  25 17 
n
n
 1
2 
s 
2
 n1  1 s12   n2  1 s22
n1  n2  2
 The t value from sample (3.1943) is greater than the critical value (2.0211).
We reject the null hypothesis (accept the alternative hypothesis) in other
words, this month Monterey and Marina have different mean temperature.
Method: 2
assume 1   2

Identifying a critical value with a degree-of-freedom of
 s12 s22 
  
 n1 n2 
2
2
2
 s12 
 s22 
 
 
 n1    n2 
n1  1 n2  1
2
6
 4
  
 25 17 
2
  4 2  6 2 
 
  
  25    17  
 26
18 




 2  31.2837  31
1
9 * t0.025,30  1* t0.025,40   2.0399
10
 Compute the t value from the sample
x  x2
75.5  73.3
t 1

 3.0718
4 / 25  6 /17
s12 s22

n1 n2
t0.025,31 
 The t value from sample (3.0718) is greater than the critical value (2.0399).
We reject the null hypothesis (accept the alternative hypothesis) in other
words, this month Monterey and Marina have different mean temperature