An investment analyst takes a random sample of 100 aggressive equity funds and calculates the average beta as 1.7. The sample betas have a standard deviation of 0.4. Using a 95% confidence interval and a z-statistic, which of the following statements about the confidence interval and its interpretation is most likely accurate? The analyst can be confident at the 95% level that the interval: A) 1.580 to 1.820 includes the mean of the population beta.
B) 1.622 to 1.778 includes the mean of the population beta.
C) 1.634 to 1.766 includes the mean of the population beta.

B) 1.622 to 1.778 includes the mean of the population beta.

Step-by-step explanation:

We have the standard deviation for the sample, so the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom,which is the sample size subtracted by 1. So

df = 100 - 1 = 99

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.9842

The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1.7 - 0.078 = 1.622.

The upper end of the interval is the sample mean added to M. So it is 1.7 + 0.078 = 1.778.

Thus the correct answer is given by option B.

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.