Sample of 80 people randomly asked in casino shows that average lost per person is $160 with a standard deviation ó = $40. Based on this data create a 99% confidence interval for population mean. Does loss of $180 for single person contradict expectation? Explain your answer.
Formula:
CI99 = mean + or - 2.575 (sd/√n)
Substitute what is given into the formula:
CI99 = 160 + or - 2.575 (40/√80)
I'll let you take it from here. Once you have the interval calculated, you should be able to answer the question in the problem.
I hope this will help get you started.
What does "or" mean?
To create a 99% confidence interval for the population mean, we can use the formula:
Confidence Interval = sample mean ± (critical value * standard deviation / √sample size)
First, let's calculate the critical value. Since we want a 99% confidence interval, we need to find the z-score that corresponds to a confidence level of 99%. The z-score for a 99% confidence level is approximately 2.576.
Next, let's input the values we have:
Sample mean = $160
Standard deviation (σ) = $40
Sample size (n) = 80
Now we can calculate the confidence interval:
Confidence Interval = $160 ± (2.576 * $40 / √80)
= $160 ± (2.576 * $40 / 8.94)
= $160 ± $11.46
Therefore, the 99% confidence interval for the population mean is $148.54 to $171.46.
Now, let's analyze whether a loss of $180 for a single person contradicts the expectation. In this case, we are given:
Loss for a single person = $180
If we compare $180 to the confidence interval we calculated, we can see that $180 falls outside the confidence interval. This means that a loss of $180 is higher than the upper bound of the confidence interval.
Since the confidence interval represents a range within which we are 99% confident that the population mean falls, a loss of $180 contradicts the expectation based on the sample data. This suggests that a single person's loss of $180 is higher than what is expected based on the findings from the sample of 80 people in the casino.