Unit Test — Unit test Answers

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A computer company wants to determine if there is a difference in the proportion of defective computer chips in a day’s production from two different production plants, A and B. A quality control specialist takes a random sample of 100 chips from the first hour of production from plant A and determines that there are 12 defective chips. The specialist then takes a random sample of 100 chips from the last hour of production from plant B and determines that there are 10 defective chips. Assuming conditions for inference are met, what is the 90% confidence interval for the true difference in proportions of defective chips from a day’s production between the two plants? Find the z-table here.

A
(0.12 minus 0.10) plus-or-minus 1.96 StartRoot StartFraction 0.12 (1 minus 0.12) Over 100 EndFraction + StartFraction 0.10 (1 minus 0.10) Over 100 EndFraction EndRoot
Option A
B
(0.12 minus 0.10) plus-or-minus 1.65 StartRoot StartFraction 0.12 (1 minus 0.12) Over 100 EndFraction + StartFraction 0.10 (1 minus 0.10) Over 100 EndFraction EndRoot
Option B
C
(0.88 minus 0.90) plus-or-minus 1.96 StartRoot StartFraction 0.88 (1 minus 0.88) Over 100 EndFraction + StartFraction 0.90 (1 minus 0.90) Over 100 EndFraction EndRoot
Option C
D
(0.88 minus 0.90) plus-or-minus 1.65 StartRoot StartFraction 0.88 (1 minus 0.88) Over 100 EndFraction + StartFraction 0.90 (1 minus 0.90) Over 100 EndFraction EndRoot
Option D
24

A student believes that a certain number cube is unfair and is more likely to land with a six facing up. The student rolls the number cube 45 times and the cube lands with a six facing up 12 times. Assuming the conditions for inference have been met, what is the 99% confidence interval for the true proportion of times the number cube would land with a six facing up?

A
0.27 plus-or-minus 2.58 StartRoot StartFraction 0.27 (1 minus 0.27) Over 45 EndFraction EndRoot
Option A
B
0.73 plus-or-minus 2.33 StartRoot StartFraction 0.73 (1 minus 0.73) Over 45 EndFraction EndRoot
Option B
C
0.27 plus-or-minus 2.33 StartRoot StartFraction 0.27 (1 minus 0.27) Over 45 EndFraction EndRoot
Option C
D
0.73 plus-or-minus 2.58 StartRoot StartFraction 0.73 (1 minus 0.73) Over 45 EndFraction EndRoot
Option D
25

A newspaper poll found that 54% of the respondents in a random sample of voters in the city plan to vote for candidate Roberts. A 95 percent confidence interval for the population proportion is 0.54 ± 0.06. Based on this interval, what can the newspaper report?

A
There is a 95% chance that Roberts will win.
B
The race is too close to call, so no prediction about who will win should be made.
C
With 95% confidence, there is convincing evidence that Roberts will win.
D
The poll predicts Roberts will win, but there is a 5% chance that the prediction is incorrect due to sampling error.

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