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1.State the claim mathematically. Then write the null and alternative hypothesis.
Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed.
a. A research hospital claims that more than 3.7% of the population suffers from high
blood pressure. Use p, the represent the population proportion.
Answer:
Claim: The claim “that more than 3.7% of the population suffers from high blood
pressure”
H0 : p = 0.037
Ha : p > 0.037
Test : Because Ha contains “greater than 0.037”, the hypothesis test is right-tailed.
b.The highest acceptable level of pesticide found in quail has been limited to 0.35 parts
per million (ppm). A hypothesis test is perform to test if the level of the pesticide is
higher.
Answer:
Claim: The mean level of the pesticide found in quail is higher than 0.35 parts per
million.
H0 : μ = 0.35
Ha : μ > 0.35
Test : Because Ha contains “greater than 0. 35”, the hypothesis test is right-tailed.
2.Use the method specified to perform the hypothesis test for the population
mean μ. WeatherBug say that the mean daily high for December in a large Florida city is
76(degrees) F. WFLA weather suspects that this temperature is not accurate. A
hypothesis test is performed to determine if the mean is actually lower than 76(degrees)
F. Assume that the population standard deviation of σ = 5.6(degrees) F. A sample of
mean daily temperatures for December over the past 40 years gives x=74 F. At a = 0.01,
does the data provide sufficient evidence to conclude that the mean temperature is lower
than 76 F.
Use the critical value z0 method from the normal distribution.
Answer
Let μ be the mean daily high temperature for December in a large Florida city.
1. H0 : μ = 76
Ha : μ < 76
2. α = 0.01
3.Test statistic:
Since the population standard deviation σ is known, the test statistic for testing H0 is
z = (x- 76)/ (σ/Sqrt(n)) follows a Standard normal distribution
Here it is given that xbar = 74, σ = 5.6 and n =40.
Thus, the test statistic,
z = (x- 76)/ (σ/Sqrt(n))
= (74-76)/(5.6/Sqrt(40))
= -2/0.8854
= - 2.2588
4.P-value or critical z0 or t0.
Since α = 0.01, the critical value z0 is given by
Critical z0 = -2.33
5.Rejection Region:
Reject H0 if z < -2.33
6.Decision:
Since z = -2.2588 > -2.33, we do not reject the null hypothesis H0.
7.Interpretation:
The data do not provide sufficient evidence to conclude that the mean temperature
is lower than 76 F at 1% level of significance.
b.Use the P-value method
Answer
1. H0 : μ = 76
Ha : μ < 76
2. α = 0.01
3.Test statistic:
Since the population standard deviation σ is known, the test statistic for testing H0 is
z = (x- 76)/ (σ/Sqrt(n)) follows a Standard normal distribution
Here it is given that xbar = 74, σ = 5.6 and n =40.
Thus, the test statistic,
z = (x- 76)/ (σ/Sqrt(n))
= (74-76)/(5.6/Sqrt(40))
= -2/0.8854
= - 2.2588
4.P-value or critical z0 or t0.
P-value = P[Z < -2.2588] = 0.0119
5.Rejection Region:
Reject H0 if P-value < 0.01
6.Interpretation:
Since the P-value = 0.0119 > 0.01, we do not reject the null hypothesis H0. Thus
the data do not provide sufficient evidence to conclude that the mean temperature is
lower than 76 F at 1% level of significance.
3. A local tire store suspects that the mean life of a new discount tire is less that 39,000
miles. To check the claim, the store selects randomly 18 of these new discount tires.
When they are tested, it is found that the mean life is 38,250 miles with a sample standard
deviation s = 1200 miles.
a. Use the critical value z0 method from the normal distribution to test for the population
mean μ. Test the company’s claim at the level of significance a = 0.05.
Answer
Let μ be the mean life of a new discount tire.
1. H0 : μ = 39,000
Ha : μ < 39,000
2. α = 0.05
3.Test statistics:
t = (xbar- 39,000)/ (s/Sqrt(n))
Here it is given that xbar = 38,250, s = 1200 and n =18.
Thus, the test statistic,
t = (xbar- 39,000)/ (s/Sqrt(n))
= (38,250 – 39,000)/(1200/Sqrt(18))
= -750/282.8427
= - 2.6517
4.P-value or critical z0 or t0.
Since α = 0.05, from standard normal table, the critical value z0 is given by,
Critical z0= -1.645
5.Rejection Region:
Reject H0 if t < -1.645
6.Decision:
Since t = - 2.6517 < -1.645, we reject the null hypothesis H0.
7.Interpretation:
The mean life of a new discount tire is less than 39,000 miles at 5% level of
significance.
b. Use the critical value z0 method from the normal distribution to test for the population
mean U. Test the company’s claim at the level of significance A = 0.01
Answer
1. H0 : μ = 39,000
Ha : μ < 39,000
2. α = 0.01
3.Test statistics:
The test statistic for testing H0 is
t = (xbar- 39,000)/ (s/Sqrt(n))
Here it is given that xbar = 38,250, s = 1200 and n =18.
Thus, the test statistic,
t = (xbar- 39,000)/ (s/Sqrt(n))
= (38,250 – 39,000)/(1200/Sqrt(18))
= -750/282.8427
= - 2.6517
4.P-value or critical z0 or t0.
Since α = 0.01, from standard normal distribution, the critical value z0 is given
by,
Critical z0= -2.33
5.Rejection Region:
Reject H0 if t < -2.33
6.Decision:
Since t = - 2.6517 < -2.33, we reject the null hypothesis H0.
7.Interpretation:
The mean life of a new discount tire is less than 39,000 miles at 1% level of
significance.
[Usually if the population standard deviation is unknown and the sample size is small
(n<30) usually t-test is used to test the significance of the population mean. In the given
question it is requested to use the critical value z0 method from the normal distribution to
test for the population mean. If the question is wrong use the t-test using student’s t
distribution. Otherwise ignore the following part. The testing procedure using Student’s t
distribution is given below.
a)
1. H0 : μ = 39,000
Ha : μ < 39,000
2. α = 0.05
3.Test statistics:
Since the population standard deviation is unknown, the test statistic for testing H0 is
t = (xbar- 39,000)/ (s/Sqrt(n)) follows a Student’s t distribution with (n-1) degrees
of freedom.
Here it is given that xbar = 38,250, s = 1200 and n =18.
Thus, the test statistic,
t = (xbar- 39,000)/ (s/Sqrt(n))
= (38,250 – 39,000)/(1200/Sqrt(18))
= -750/282.8427
= - 2.6517
4.P-value or critical z0 or t0.
Since α = 0.05, from Student’s t table with (n-1) =(18-1) =17 degrees of freedom,
the critical value t0 is given by,
Critical t0= -1.7396
5.Rejection Region:
Reject H0 if t < -1.7396
6.Decision:
Since t = - 2.6517 < -1.7396, we reject the null hypothesis H0.
7.Interpretation:
The mean life of a new discount tire is less than 39,000 miles at 5% level of
significance.
b)
1. H0 : μ = 39,000
Ha : μ < 39,000
2. α = 0.01
3.Test statistics:
Since the population standard deviation is unknown, the test statistic for testing H0 is
t = (xbar- 39,000)/ (s/Sqrt(n)) follows a Student’s t distribution with (n-1) degrees
of freedom.
Here it is given that xbar = 38,250, s = 1200 and n =18.
Thus, the test statistic,
t = (xbar- 39,000)/ (s/Sqrt(n))
= (38,250 – 39,000)/(1200/Sqrt(18))
= -750/282.8427
= - 2.6517
4.P-value or critical z0 or t0.
Since α = 0.01, from Student’s t table with (n-1) =(18-1) =17 degrees of freedom,
the critical value t0 is given by,
Critical t0= -2.5669
5.Rejection Region:
Reject H0 if t < -2.5669
6.Decision:
Since t = - 2.6517 < -2.5669, we reject the null hypothesis H0.
7.Interpretation:
The mean life of a new discount tire is less than 39,000 miles at 1% level of
significance.]
A flash drive manufacturer has set a standard on their production process. When defects
exceed 3%, the production process is unacceptable. A random sample of 85 drives is
tested. The defective rate is 5.9%. Use a level of significance of a = 0.01 to test to see if
you have sufficient evidence to support the claim that the defective rate exceeds 3%.
Let p denote the defective rate.
1. H0 : p = 3%
Ha : p > 3%
2. α = 0.01
3.Test statistics:
The test statistic for testing H0 is
z = (p^ - p0)/Sqrt[p0(1-p0)/n), follows a standard normal distribution.
Here it is given that, p^ = 0.059, n =85, p0 =0.03
Thus the test statistic is given by,
z = (p^ - p0)/Sqrt[p0(1-p0)/n]
= (0.059 – 0.03)/Sqrt[0.03*0.97/85]
= 0.029/0.0185
= 1.5673
4.P-value or critical z0 or t0.
Since α = 0.01, the critical value z0 is given by
Critical z0 = 2.33
5.Rejection Region:
Reject H0 if z > 2.33
6.Decision:
Since z = 1.5673 < 2.33, we do not reject the null hypothesis H0
7.Interpretation:
There is no sufficient evidence to support the claim that the defective rate exceeds
3% at 0.01 significance level.