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7. A sample size of 45 is used to test H0: =75 vs. Ha: < 75.
Given that x-bar= 72.9 and σ = 4.3, answer the following questions (4 pts)
a. What is the computed value of the test statistic?
b. What distribution does the test statistic have when the null hypothesis is true?
c. Is the alternative hypothesis one-tailed or two-tailed?
d. What is the p-value?
Answer:
Here it is given that, n = 45, x-bar = 72.9 and σ = 4.3.
a) The test statistic for testing the null hypothesis is
z = (x-bar – 75)/( σ/√n)
= (72.9 – 75)/(4.3/√45)
= -3.2761
b) When the null hypothesis is true the test statistic follows a Standard Normal
distribution.
c) Here the alternative hypothesis is one-tailed.
d) The p-value of the test is given by,
p-value = P[Z < -3.2761]
= 0.0005
8. List the steps involved in hypothesis testing using: a) the traditional/classical
method; b) the probability-value method. Provide an example of each.
Steps involved in hypothesis testing using: a) the traditional/classical method
The first step in hypothesis testing is to specify the null hypothesis (H0) and the
alternative hypothesis (H1).
The second step is to specify the α level also known as the significance level. Typical
values are 0.05 and 0.01.
The third step is to calculate a statistic analogous to the parameter specified by the null
hypothesis.
The fourth step is to specify the critical region or rejection region of the test. The critical
region is the set of those values of the test statistic which leads to the rejection of the null
hypothesis. The values that mark the boundaries of the critical region are called critical
values. The critical values of the critical region are usually obtained using the distribution
of the test statistic such that the significance level is α.
Next, if the test statistic computed in Step 3 is in the critical region obtained in Step 4,
then the null hypothesis is rejected; otherwise the null hypothesis is not rejected. When
the null hypothesis is rejected, the outcome is said to be "statistically significant" when
the null hypothesis is not rejected then the outcome is said be "not statistically
significant."
Steps involved in hypothesis testing using the probability-value method
The first step in hypothesis testing is to specify the null hypothesis (H0) and the
alternative hypothesis (H1).
The second step is to specify the α level also known as the significance level. Typical
values are 0.05 and 0.01.
The third step is to calculate a statistic analogous to the parameter specified by the null
hypothesis.
The fourth step is to calculate the probability value (often called the p value). The p value
is the probability of obtaining a statistic as different or more different from the parameter
specified in the null hypothesis as the statistic computed from the data. The calculations
are made assuming that the null hypothesis is true.
Next, the probability value computed in Step 4 is compared with the significance level
chosen in Step 2. If the p value is less than or equal to the significance level, then the null
hypothesis is rejected; if the probability is greater than the significance level then the null
hypothesis is not rejected.
Note that the main difference in the traditional/classical method and the probability-value
method are in Steps 4 and 5.
In general the different steps involved in hypothesis testing using: a) the
traditional/classical method; b) the probability-value method can be summarized as
follows
1. Select your hypothesis
2. Select a significance level
3. Find the test statistic
4. Find the p-value or determine the critical region.
5. Draw your conclusions.
We shall discuss the following as an example showing the steps involved in hypothesis
testing using: a) the traditional/classical method and b) the probability-value method.
A researcher claims that the average age of people who buy lottery tickets is 70. A
sample of 30 is selected and their ages are recorded as shown below. The standard
deviation is 16. At alpha = 0.05 is there enough evidence to reject the researcher’s
claim?
49 63 90 52 22 80 72 56 82 56
24 46 70 74 70 61 65 71 39 74
79 76 71 49 62 68 71 67 69 45
The steps involved in hypothesis testing using traditional/classical method is given below

Step 1: Formulation of the null and the alternative hypotheses
Let  denote the average age of people who buy lottery tickets.
Here we want to test the null hypothesis H 0 :   70 against the alternative
hypothesis H1 :   70

Step 2: Specification of the level of significance
Here the level of significance, α is 0.05

Step3: Calculation of the test statistic
The test statistic used to test H0 is
x  70
Z
N (0,1) , the Standard normal distribution.
/ n
From the given data, n =30, x = 62.43 (=1873/30) and  = 16
x  70 62.43  70
So Z 
= -2.59 and hence

 / n 16 / 30
Z = 2.59

Step 4: Definition of the critical region
Since alpha = 0.05, the critical region of the test is Z > 1.96 .

Step 5: Selection of the appropriate hypothesis
Since Z = 2.59> 1.96, we reject the null hypothesis. So there is enough evidence to
reject the researcher’s claim that “the average age of people who buy lottery tickets is
70”.
The steps involved in hypothesis testing using the probability-value method

Step 1: Formulation of the null and the alternative hypotheses
Let  denote the average age of people who buy lottery tickets.
Here we want to test the null hypothesis H 0 :   70 against the alternative
hypothesis H1 :   70

Step 2: Specification of the level of significance
Here the level of significance, α is 0.05

Step3: Calculation of the test statistic
The test statistic used to test H0 is
x  70
Z
N (0,1) , the Standard normal distribution.
/ n
From the given data, n =30, x = 62.43 (=1873/30) and  = 16
x  70 62.43  70
So Z 
= -2.59

 / n 16 / 30

Step 4: Calculation of the p-value
Since the alternative hypothesis is H1 :   70 , the probability value or p-value of the test
is given by
p-value = P  Z > 2.59 = 0.0096

Step 5: Selection of the appropriate hypothesis
Since the p-value = 0.0096 is less than the significance level, α = 0.05, we reject the null
hypothesis. So there is enough evidence to reject the researcher’s claim that “the average
age of people who buy lottery tickets is 70”.
1. An estimator is consistent if, as the sample size decreases, the value of the estimator
approaches the value of the parameter estimated. __F__ T/F
Answer: False
Explanation: An estimator is consistent if, as the sample size increases, the value of the
estimator approaches the value of the parameter estimated.
2. For a specific confidence interval, the larger the sample size, the smaller the maximum
error of estimate will be. __T___ T/F
Answer: True
3. When we reject the null hypothesis, we are certain that the null hypothesis is false.
__F_ T/F
Answer: False
Explanation: When we reject the null hypothesis, we may conclude that the null
hypothesis is false at certain level of significance based on the sample information. But
there is a chance that the decision may be wrong.
4. The alternative hypothesis, sometimes referred to as the research hypothesis, is
supported by using the sample evidence to contradict the null hypothesis. _T_ T/F
Answer: True
5.  is the measure of the area under the curve of the standard score that lies in the
rejection region for the null hypothesis. _T_ T/F
Answer: True
6. Rejection of a null hypothesis that is false is a Type II error. _F_ T/F
Answer: False
Explanation: Acceptance of a null hypothesis that is false is a Type II error.
7. The maximum error of estimate is controlled by three factors: level of confidence,
sample size, and standard deviation. __T__ T/F
Answer: True
8. You are constructing a 95% confidence interval using the information: n = 50, x-bar=
54.3, s = 2.1 and E = 0.65. What is the value of the middle of the interval?
A. 2.1
B. 54.3
C. 50
D. 0.95
Answer: B. 54.3
Explanation: The value of the middle of the interval is x-bar = 54.3.
9. Which of the following would be the correct hypotheses for testing the claim that the
mean life of a battery for a cellular phone (while the phone is left on) is at least 24 hours?
A.
B.
C.
D.
H0:   24 and Ha:  < 24
H0:  = 24 and Ha:   24
H0:   24 vs. Ha:  > 24
H0:  > 24 vs. Ha:   24
Answer: C. H0:   24 vs. Ha:  > 24
Explanation: The alternative hypothesis should be the mean life of a battery for a cellular
phone () is at least 24 hours ( > 24).
10. Which of the following would be the alternative hypothesis in testing the claim that
the mean distance students commute to campus is no less than 8.2 miles?
A. Ha:   8.2
B. Ha:  > 8.2
C. Ha:  < 8.2
D. Ha:   8.2
Answer: B. Ha:  > 8.2
Explanation: The alternative hypothesis should be the mean distance students commute to
campus () is no less than 8.2 miles ( > 8.2).
Part II. Short Answers & Computational Questions
1. A study was conducted to estimate the mean amount spent on birthday gifts for a
typical family having two children. A sample of 175 was taken, and the mean amount
spent was $250. Assuming a standard deviation equal to $45, find the 95% confidence
interval for ?, the mean for all such families. (4 pts)
If x-bar denotes the sample mean and σ denotes the standard deviation, the 95%
confidence interval for  is (x-bar – 1.96 * SE, x-bar + 1.96 * SE), where SE = σ/n
Here it is given that
x-bar = 250, σ = 45 and n = 175
Therefore, SE = σ/n = 45/175 = 3.4017
The 95% CI for the mean amount spent is given by
(x-bar – 1.96 * SE, x-bar + 1.96 * SE)
= (250 – 1.96 * 3.4017, 250 + 1.96 * 3.4017)
= (250 – 6.67, 250 + 6.67)
= ($243.33, $256.67)
2. What sample size would be needed to estimate the population mean to within one-third
standard deviation with 95% confidence? (4 pts)
If σ denotes the standard deviation, the margin of error is given as σ/3
Thus the sample size needed to estimate the population mean to within one-third standard
deviation with 95% confidence is
n = (z * σ /Margin of error)^2 = [1.96 * σ /( σ/3)]^2 = (1.96 * 3)^2 = 34.57
Hence a sample size of 35 is required.
3. A 95% confidence interval estimate for a population mean was computed to be (63.4 to
68.2). Determine the mean of the sample, which was used to determine the interval
estimate. (4 pts)
Mean = (63.4 + 68.2)/2 = 131.6/2 = 65.8
Thus the mean of the sample = 65.8
4. In testing the hypothesis, H0:  31.5 and Ha:  < 31.5, using the p-value approach, a
p-value of 0.0409 was obtained. If  = 8.7, find the sample mean which produced this pvalue given that the sample of size n = 50 was randomly selected. (4 pts)
The z- score corresponding to a p- value (Left-tailed test) of 0.0409 is -1.74
We have, the z score, z= (x-bar - )/SE, where SE = σ/n
Therefore,
x-bar =  + z * SE
= 31.5 + (-1.74)(8.7/50)
= 31.5 – 2.14
= 29.36
5. To test the null hypothesis that the average lifetime for a particular brand of bulb is 725
hours versus the alternative that the average lifetime is different from 725 hours, a sample
of 100 bulbs is used. If the standard deviation is 50 hours and  is equal to 0.01, what
values for x-bar will result in rejection of the null hypothesis. (4 pts)
Here it is given that,
σ = 50, n = 100
Therefore,
SE = σ/n = 50/100 = 5
The critical value of z corresponding to  = 0.01 (Two-tailed) is 2.576
The test statistic for testing the null hypothesis is
z = (x-bar – )/SE
We reject the null hypothesis if z < -2.576 or z > 2.576
To reject the null hypothesis, we need (x-bar – 725)/5 < -2.576 or (x-bar – 725)/5 > 2.576
x-bar < 725 - 5 * 2.576 or x-bar > 725 + 5 * 2.576
x-bar < 725 – 12.88 or x-bar > 725 + 12.88
x-bar < 712.12 hours or x-bar > 737.88 hours
6. Describe the action that would result in a type I error and a type II error if each of the
following null hypotheses were tested. (4 pts)
a. H0: There is no waste in US Defense Department spending
b. H0: This fast-food menu is not low fat
(a) If there really is no waste in US Defense Department spending and if H0 is rejected
then a Type I error is committed. If there is waste in US Defense Department spending
and if we fail to reject H0 then a Type II error is committed.
(b) If the fast food menu is really not low fat and if H0 is rejected then a Type I error is
committed. If the fast food menu is low fat and if we fail to reject H0 then a Type II error
is committed.
7. By measuring the amount of time it takes a component of a product to move from one
workstation to the next, an engineer has estimated that the standard deviation is 5.1
seconds. (4 pts)
a. How many measurements should be made in order to be 95% certain that the
maximum error of estimation will not exceed 1.5 seconds?
b. What sample size is required for a maximum error of 2.5 seconds?
(a) Here it is given that, the margin of error = 1.5 and standard deviation, σ = 5.1
Thus the number of measurements should be made in order to be 95% certain that the
maximum error of estimation will not exceed 1.5 seconds is
n = (z * σ /Margin of error)^2 = (1.96 * 5.1/1.5)^2 = 44.41
Thus a sample size of 45 is required.
(a) Here it is given that, the margin of error = 2.5
Thus the sample size required for a maximum error of 2.5 seconds is
n = (z * σ /Margin of error)^2 = (1.96 * 5.1/2.5)^2 = 15.99
Thus a sample size of 16 is required.
8. Determine the critical region and critical values for z that would be used to test the null
hypothesis at the given level of significance, as described in each of the following: (6pts)
a. H0:   51 and Ha:  > 51,  = 0.10
b. H0:   28 and Ha:  < 28,  = 0.01
c. H0:  = 93 and Ha:   93,  = 0.05
a) Since the alternative hypothesis is Ha:  > 51, the critical region is the region in the
right tail (upper tail). At  = 0.10, the critical value is z = 1.28.
Thus the critical region is: Reject H0 if the calculated value of the test statistic > 1.28
b) Since the alternative hypothesis is Ha:  < 28, the critical region is the region in the left
tail (lower tail). At  = 0.01, the critical value is z = -2.33.
Thus the critical region is: Reject H0 if the calculated value of the test statistic < -2.33
c) Since the alternative hypothesis is Ha:   93, the critical region is the region in the two
tails. At  = 0.05, the critical value is z = ± 1.96.
Thus the critical region is: Reject H0 if the calculated value of the test statistic < -1.96 or
if the calculated value of the test statistic > 1.96