Point and Interval Estimates & Hypothesis Testing - Problems solution

      1.            The following table shows the order size in dollars of 30 home shoppers.

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75   109   32    54    121    80     96   47    67    115 

29    70    89   100     48    40   137   75    39      88 

99   140   112    87   122   75     54    92   89    153

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a.       Construct a 95 percent confidence interval around the mean.

b.      Compute the point estimate for the mean of the population

Solution:

a)

For n= 30 , the sample acts as Normal distribution (central limit Theorem)

n ≥ 30

Using Minitab

Variable	N	N*	Mean	SE Mean	StDev	Min	Q1	Median	Q3	Max
C1	30	0	84.47	6.02	32.98	29	54	87.5	109.75	153

The Sample behaves as normally distributed, the population standard deviation (σ) is not known and the sample size is large (n ≥ 30) so I must use z-distribution.

n = 30

x̅ = 84.47

s = 32.98

z_α/2 = 1.96         for 95% confidence level

σ_x̅  = s/_n^0.5 =  6.021

•  Upper limit = x+ z_α/2 σ_x̅

• Lower limit    = x - z_α/2 σ_x̅

•  Upper Limit = x̅ + 1.96 σ_x̅  = 84.47+ 1.96  x 6.021 =  96.271  $  

 •  Lower Limit    = x̅ - 1.96 σ_x̅  = 84.47 - 1.96 x 6.021 = 72.668  $

Φ ( (  x- z_α/2 σ_x̅) < x < ( x+ z_α/2 σ_x̅ ) , x, σ_x̅) =1 - α

Φ (72.668< x < 92.271) = 1-0.05=0.95

b)  The point estimate for the mean of the population: 

       x̅ = 84.47

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       2.            The following sample of size 11 was taken from a population that is normally distributed:

121    136    102    115    126    106    115    132    125    108    130

Construct a 98 percent confidence interval around the mean

Solution:

Using Minitab

Variable	N	N*	Mean	SE Mean	StDev	Min	Q1	Median	Q3	Max
C1	11	0	119.64	3.4	11.29	102	108	121	130	136

The population is normally distributed, the population standard deviation (σ) is not known and the sample size is small (n < 30) so I must use t-distribution.

n = 11       

DF = n-1 = 10          

x̅ = 119.64

s = 11.29

t_α/2 = 2.764        for 98% confidence level 

σ_x̅  = s/_n^0.5  = 3.404

•     Upper limit = + t_α/2 σ_x̅

 •     Lower limit    =  t_α/2 σ_x̅

 •      Upper Limit = x̅ + 2.764 σ_x̅  = 119.64 + 2.764 x 3.404 =  129.048

 •      Lower Limit    = x̅ - 2.764 σ_x̅  = 119.64 - 2.764 x 3.404 = 110.231

        Φ ( (  - t_α/2 σ_x̅) < x < ( + t_α/2 σ_x̅ ) , , σ_x̅) =1 - α

        Φ (110.231< x < 129.048) = 1-0.02=0.98

  

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       3.            A student organization at a small business college claims that the average class size is greater than 35 students. Test this claim at α = 0.02, using the following sample of class size:

42    28    36    47    35    41    33    30    39    48

Assume the population is normally distributed and that σ is unknown.

Solution:

The Hypotheses are:

H₀  :   µ = 35

H₁  :   µ > 35

Variable	N	N*	Mean	SE Mean	StDev	Min	Q1	Median	Q3	Max
C1	10	0	37.9	2.13	6.74	28	32.25	37.5	43.25	48

The population is normally distributed, the population standard deviation (σ) is not known and the sample size is small (n < 30) so I must use t-distribution.

DF = n-1 = 9

x̅ = 37.9

s = 6.74

t_α =  2.398   for 98% confidence level ( α = 0.02 )      - One tail test

σ_x̅  =s/_n^0.5 = 2.1313

The acceptance interval for H₀ is given by:

P (  –∞ < t < t_α  ) =1- α

P (  –∞ < t < 2.398  ) =0.98

Then t₀ < t _α    , and in the H₀ Acceptance Interval so we do not reject H_0 , and can say that  the average class size is 35 students.

  

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      4.            The tensile strength of a fiber used in manufacturing cloth is of interest to the purchaser.  Previous experience indicates that the standard deviation of tensile strength is 2 psi. A random sample of eight fiber specimens is selected, and the average tensile strength is found to be 127 psi.

a.       Test the hypothesis that the mean tensile strength equals 125 psi versus the alternative that the mean exceeds 125 psi. Use a = 0.05

b.       What is the P-value for this test?

c.       Discuss why a one-sided alternative was chosen in part (a).

d.       Construct a 95% lower confidence interval on the mean tensile strength.

Solution:

σ =2 Psi

x̅ = 127

n=8

a)

The Hypotheses are:

H₀  :   µ₀ = 125

H₁  :   µ₁ > 125

This will be One tail test

          α = 0.05

The population standard deviation (σ) is known and the sample size is small (n < 30) so I must use Z-distribution.

z_α =  1.645    for 95% confidence level ( α = 0.05 )      - One tail test

The acceptance interval for H₀ is given by:

 P (  –∞ < z < z_α  ) =1- α

P (  –∞ < z < 1.645) =0.95

Then z₀ > z _α    , and in the H₀ rejection Interval so we reject H_0, and accept H₁ (the mean greater than 125 psi) .

 b)

The P-value would be computed using only the area in the upper tail of the standard normal distribution, because the alternative hypothesis is one-sided.

P = 1- Φ(Z₀) =  1- Φ (2.8284)  = 1- 0.9977 = 0.002339

c)

 one-sided alternative was chosen because the experimenter seeks to have an average tensile strength greater than in the null Hypothesis.

d)

z_α =  1.645    for 95% confidence level ( α = 0.05 )      - One tail test

• Lower limit    =  x̅   - z_α σ_x̅

•  Lower Limit = x̅ - 1.645 σ_x̅  = 127 - 1.645 x 0.7071= 125.836

x̅ ≥ 125.836               ,   x̅ Should be grated than 125.836

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Other Article :

Experiments with a Single Factor - Problems solution