Question

You play a game and win 136 out of 270 times. Find a 95% confidence interval for the probability of winning the game.

Answer 1

The 95% confidence interval for the probability of winning the game is calculated using the sample proportion, standard error, z-score for a 95% confidence level, and margin of error. It is found to be approximately (0.4434, 0.5646).

Step-by-step explanation:

To find a 95% confidence interval for the probability of winning the game, we can use the proportion of times the student won the game as an estimate for the probability of winning, along with the standard error for a proportion. The student won 136 out of 270 games, so the sample proportion (p-hat) is calculated as 136/270 which approximately equals 0.504. The standard error (SE) for the proportion is calculated using the formula SE = √(p-hat * (1 - p-hat) / n), where n is the number of trials.

We also need to determine the z-score associated with a 95% confidence level, which places 2.5% probability in each tail of the distribution. For a 95% confidence interval, the z-score is approximately 1.96. The margin of error (MOE) can be computed as: MOE = z-score * SE, and finally, we can find the confidence interval with: (p-hat - MOE, p-hat + MOE).

Step-by-Step Confidence Interval Calculation:

1. Calculate sample proportion (p-hat) = 136/270 ≈ 0.504.
2. Calculate the standard error: SE = √(0.504 * (1 - 0.504) / 270) ≈ 0.0309.
3. Find the z-score for a 95% confidence level: z-score ≈ 1.96.
4. Calculate the margin of error: MOE = 1.96 * 0.0309 ≈ 0.0606.
5. Calculate the confidence interval: (0.504 - 0.0606, 0.504 + 0.0606) = (0.4434, 0.5646).

Thus, the 95% confidence interval for the probability of winning the game is approximately (0.4434, 0.5646).

Answer 2

To find a 95% confidence interval for the probability of winning the game, you can use the formula CI = p ± Z * sqrt((p(1-p))/n), where p is the sample proportion, Z is the z-score for a 95% confidence level, and n is the sample size.

Step-by-step explanation:

To find a 95% confidence interval for the probability of winning the game, we can use the formula:

CI = p ± Z * sqrt((p(1-p))/n)

Where:

• CI is the confidence interval
• p is the sample proportion (136/270)
• Z is the z-score for a 95% confidence level, which is approximately 1.96
• sqrt is the square root function
• n is the sample size (270)

Plugging in the values, we have:

CI = (136/270) ± 1.96 * sqrt(((136/270)(1-(136/270)))/270)

Calculating this expression gives us the 95% confidence interval for the probability of winning the game.