CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTINGpublic.csusm.edu/ckumar/BUS 304 course material/course...
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CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING 8.1. a. 20 : 20 : > ≤ μ μ A o H H b. 50 : 50 : ≠ = μ μ A o H H c. 35 : 35 : < ≥ μ μ A o H H d. 87 : 87 : > ≤ μ μ A o H H e 6 : 6 : > ≤ μ μ A o H H 8.3. a. If x > 205.2344 reject H o If x < 205.2344 do not reject H o x α = 200 + 1.645(45/ 200 ); x α = 205.2344 If z > 1.645 reject H o If z < 1.645 do not reject H o b. z = (204.50 – 200)/(45/ 200 ) = 1.41; Since 1.41 < 1.645 do not reject H o Since 204.5 < 205.2344 do not reject H o c. The alternative hypothesis. The burden of proof is always to on the alternative hypothesis. 8.4. a. If z > 1.88 reject H o If z < 1.88 do not reject H o If p-value < 0.03 reject H o b. z = (24.85 – 24.78)/(9/ 50 ) = 0.0550; Since 0.0550 < 1.88 do not reject H o p-value = 0.4781
Transcript of CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTINGpublic.csusm.edu/ckumar/BUS 304 course material/course...
CHAPTER 8
INTRODUCTION TO HYPOTHESIS TESTING
8.1. a.
20:20:
>≤
μμ
A
o
HH
b.
50:50:
≠=
μμ
A
o
HH
c.
35:35:
<≥
μμ
A
o
HH
d.
87:87:
>≤
μμ
A
o
HH
e
6:6:
>≤
μμ
A
o
HH
8.3.
a. If x > 205.2344 reject Ho If x < 205.2344 do not reject Ho
x α = 200 + 1.645(45/ 200 ); x α = 205.2344
If z > 1.645 reject Ho If z < 1.645 do not reject Ho
b. z = (204.50 – 200)/(45/ 200 ) = 1.41; Since 1.41 < 1.645 do not reject Ho
Since 204.5 < 205.2344 do not reject Ho c. The alternative hypothesis. The burden of proof is always to on the alternative hypothesis.
8.4.
a. If z > 1.88 reject Ho If z < 1.88 do not reject Ho
If p-value < 0.03 reject Ho b. z = (24.85 – 24.78)/(9/ 50 ) = 0.0550; Since 0.0550 < 1.88 do not reject Ho
p-value = 0.4781
Instructor’s Solutions Manual
265 8.5.
a. Even though the population standard deviation is unknown, since n = 100 is large, we can use the standard normal distribution to obtain the critical value.
x α = 4,000 - 1.645(205/ 100 ); x α = 3966.2775
If x < 3966.2775 reject Ho If x > 3966.2775 do not reject Ho
If p-value < .05, reject Ho If p-value > .05, do not reject Ho
b. Since 3980 > 3966.2775 do not reject Ho
p-value = P(z < -.9756) = .50 - .3365 = .1635 Since p-value = .1635 > 0.05 do not reject Ho c. The two research hypotheses that could have produced the null and alternative hypotheses are: The population mean is less than 4,000. The population mean is at least 4,000.
8.8.
a. If: z > 1.96 reject Ho
If: z < -1.96 reject Ho Otherwise, do not reject Ho b. z = (1,338 – 1,346)/(90/ 64 ) = -0.7111
Since z = -0.7111 > -1.96, do not reject the null hypothesis.
c. Since the null hypothesis is not rejected, a Type II error may have been committed. 8.14.
a. Ho: μ < 240 seconds Ha: μ > 240 seconds
b. Students can use Excel’s AVERAGE and STDEV functions to determine the sample mean and
standard deviation or compute them manually.
x = 219.6667 s = 57.5884
t = (219.6667 – 240)/(57.5884/ 12 ) = -1.2231
The critical t value for a one tailed test with alpha = .10 and 11 degrees of freedom is 1.3634. Since t = -1.2231 < 1.3634, do not reject the null hypothesis. There is not enough evidence to conclude that the average time exceeds 4 minutes (240 seconds).
Chapter 8: Introduction to Hypothesis Testing 266
Also could use Excel’s TDIST function to determine the p-value = 0.8766. Since p-value = 0.8766 > 0.1, do not reject Ho
c. x α = 240 + 1.3634(57.5884/ 12 ); x α = 262.6656
Since x = 219.6667 < 262.6656, do not reject the null hypothesis.
8.16.
a. Ho: μ = 24 ounces Ha: μ ≠ 24 ounces
b. t = (24.32 – 24)/(0.7/ 16 ) = 1.83
t.05/2 = + 2.1315 Since –2.1315 < 1.83 < 2.1315 do not reject Ho and conclude that the filling machine remains all right to operate.
c. Because the production control manager does not want the boxes under-filled or over-
filled. d. Using Excel’s TDIST function, p-value = 0.0872 > 0.025; therefore do not reject Ho e. Since the null hypothesis was “accepted”, a Type II error may have been committed.
