Use a 0.05 significance level to test the claim that the different samples come from populations with the same mean. Use the traditional method.
List
Reject or fail to reject?
State the conclusion as outlined in the template or book
Your hypothesis test must take up one whole page only (8 1/2″ x 11″), typed or written and must be neat.
| ANOVA | |||||
| Source of Variation | SS | df | MS | F | P-value |
| Factor | 13.500 | 3 | 4.500 | 5.171 | 0.0011 |
| Error | 13.925 | 16 | 0.870 | ||
| Total | 27.425 | 19 |
Use a 0.05 significance level to test the claim that the different samples come from populations with the same mean. Use the traditional method
Reject or fail to reject?
State the conclusion as outlined in the template or book
| ANOVA | |||||
| Source of Variation | SS | df | MS | F | P-value |
| Factor | 13.500 | 3 | 4.500 | 5.171 | 0.0011 |
| Error | 13.925 | 16 | 0.870 | ||
| Total | 27.425 | 19 |
A random sample of 760 subjects was obtained, and each was tested for left-hand writing. Results are in the table below.
| Writes with Left Hand? | ||
| Yes | No | |
| Male | 26 | 218 |
| Female | 64 | 452 |
Use a 0.05 significance level to test the claim that left-handedness is independent of gender.
Round to four decimal places if necessary.
a) Write the hypotheses.
b) The expected frequency of female who is left-handedness is
The expected frequency of male who is not left-handedness is
c) significance level: α =
d) Test statistic:
Clearly state whether test statistic in this claim is z, t, chisq. For example, “chisq=1.2345”
e) Compute p-value of the test statistic.
f) Decision: Type “yes” if reject null hypothesis. Type “no” if not to reject null hypothesis.
Use a 0.05 significance level to test the claim that pedestrian fatalities are independent of the intoxication of the driver and the intoxication of the pedestrian. Use the traditional method. Include all the expected values in the chart.
Reject or fail to reject?
State the conclusion as outlined in the template or book
| Pedestrian Intoxicated | Expected Value | Pedestrian not intoxicated | Expected Value | Totals | |
| Driver Intoxicated | 59 | 79 | |||
| Driver not Intoxicated | 266 | 581 | |||
| Totals |
Use a 0.05 significance level to test the claim that pedestrian fatalities are independent of the intoxication of the driver and the intoxication of the pedestrian. Use the traditional method. Be sure to include all the expected values in the chart.
|
Pedestrian Intoxicated |
Expected Value |
Pedestrian not intoxicated |
Expected Value |
Totals |
|
|
Driver Intoxicated |
59 |
79 |
|||
|
Driver not Intoxicated |
266 |
581 |
|||
|
Totals |
List
Reject or fail to reject?
State the conclusion
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