Download OC3140 HW/Lab 7 Hypothesis Testing

OC3140
HW/Lab 7 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 σ )
•
Start on the null hypothesis: H 0 : μ sst = 25o C
Define the alternative hypothesis: H A : μ sst > 25o C
Identify a test statistic: for a testing on means, a t distribution is used:
x − μ sst
t=
s
n
• Identifying a critical value tc with α = 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. An electrical firm manufactures light bulbs that have a length of life for grade A
is 1000 hours. A sample of 21 bulbs was test for the length of life for today
products. The sample mean and standard deviation are 990 hours and 20 hours.
Determine if today’s products must be decreased the grade?
Solution:
This is the Hypothesis Testing on the mean (see Ch. 7 p7-p8) (as small sample
and unknown σ )
•
Start on the null hypothesis: H 0 : μ = 1000hours
•
Define the alternative hypothesis: H A : μ < 1000hours
•
Identify a test statistic: for a testing on means, a t distribution is used:
x −μ
t=
s
n
• Identifying a critical value tc with α = 0.05, d . f . = n − 1 = 20 and H A is
lower one-side, we have (from the t table Ch.5 p17)
tc = −1.725 .
• For the testing sample,
n=21, mean=990, standard deviation = 20,
the t-value is
990 − 1000
t=
= −2.2913 .
20
21
• Since t ( = −2.2913) < tc ( = −1.725 ) , we reject the null hypothesis H 0 and
conclude that today’s products must be decreased the grade.
3. 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 : σ 2 ≤ 0.012
•
•
Define the alternative hypothesis: H A : σ 2 > 0.012
Identify the test statistics,
( n − 1) s 2
χ2 =
2
•
Identifying a critical value base on:
α = 0.05, d . f . = n − 1 = 49 and H a : σ 2 > 0.012 (i.e., upper one-sided),
σ
we have (from the χ 2 table Ch.5 p12),
2
χ c2 = χ 0.05,49
= 67.5 .
•
•
Compute the test statistics from the sample:
( n − 1) s 2 49 ⋅ 0.02
2
=
= 81.67 .
χ =
0.012
σ2
Since χ 2 ( = 81.67 ) > χ c2 ( = 67.5 ) , we reject H 0 and conclude that the
batteries are significantly more variable than the manufacture claimed.
4. Is the mean value of the sea surface salinity of Monterey Bay in May less than 34
psu? If the standard deviation of the salinity is 0.5 psu. The two year recode
(every day) has the mean value of 33.8 psu, respectively.
Solution:
This is the Hypothesis Testing on the mean (see Ch. 7 p7-p8) (as the large
sample with known σ )
•
Start on the null hypothesis: H 0 : μ = 34 psu
•
•
Define the alternative hypothesis: H A : μ < 34 psu
Identify a test statistic: for a testing on means, a normal distribution is
x −μ
used: z =
σ
n
• Identifying a critical value tc with α = 0.05 and H A is lower one-side, we
have (from the t table Ch.5 p17)
tc = −1.65 .
• For the testing sample,
the z-value is
33.8 − 34
z=
= −3.15 .
0.5
62
• Since z ( = −3.15 ) < zc ( = −1.65 ) , we reject the null hypothesis H 0 and
conclude the mean value of the sea surface salinity of Monterey Bay in
May is less than 34 psu.
5. The GPS receiver can receive the undulation height data with the standard
deviation of 0.5m with the good satellite signal. The data sample right now with
the size of 101 has the standard deviation of 0.55m. Determine if the satellite
signal is good (with the standard deviation of 0.5m)?
Solution:
This is the Hypothesis Testing on the Variance (see Ch. 7, p8-p9)
•
Start on the null hypothesis: H 0 : σ ≤ 0.5
•
•
Define the alternative hypothesis: H A : σ > 0.5
Identify the test statistics,
( n − 1) s 2
χ2 =
2
σ
•
Identifying a critical value base on:
α = 0.05, d . f . = n − 1 = 100 and H a : σ > 0.5 (i.e., upper one-sided),
we have (from the χ 2 table Ch.5 p12),
2
χ c2 = χ 0.05,100
= 124.342 .
•
•
Compute the test statistics from the sample:
n − 1) s 2 100 ⋅ 0.552
(
2
=
= 121 .
χ =
0.52
σ2
Since χ 2 ( = 121) < χ c2 ( = 124.342 ) , we accept H 0 and conclude that the
satellite signal is good (with the standard deviation of 0.5m).