Question

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?

0.27 plus-or-minus 2.58 StartRoot StartFraction 0.27 (1 minus 0.27) Over 45 EndFraction EndRoot
0.73 plus-or-minus 2.33 StartRoot StartFraction 0.73 (1 minus 0.73) Over 45 EndFraction EndRoot
0.27 plus-or-minus 2.33 StartRoot StartFraction 0.27 (1 minus 0.27) Over 45 EndFraction EndRoot
0.73 plus-or-minus 2.58 StartRoot StartFraction 0.73 (1 minus 0.73) Over 45 EndFraction EndRoot

Answer 1

Answer: Option #1

Explanation:

99 % confidence level - so it has to be 2.58 not 2.33 ( thats 98% confidence level )

then 12times facing u out of 45 times .... 12/45= .27

Option #1 : 0.27 +/- 2.58 StartRoot StartFraction 0.27 (1 minus 0.27) Over 45 EndFraction EndRoot

Answer 2

The 99% confidence interval for the true proportion of times the number cube would land with a six facing up is 0.27 plus-or-minus 2.58 StartRoot StartFraction 0.27 (1 minus 0.27) Over 45 EndFraction EndRoot.

What is Confidence Interval?

This is an operation which is used to measure probability that a parameter will fall between a pair of values around the mean

99 % confidence level is closer to 2.58 and 12 times facing up out of 45 times which is 12/45= .27

Hence the reason why option A was chosen as the most appropriate choice.