Question

A sample survey of 57 discount brokers showed that the mean price charged for a trade of 100 shares at $50 per share was $33.79 . The survey is conducted annually. With the historical data available, assume a known population standard deviation of $14 .

a. Using the sample data, what is the margin of error associated with a 99% confidence interval (to 2 decimals)?

b. Develop a 99% confidence interval for the mean price charged by discount brokers for a trade of 100 shares at $50 per share (to 2 decimals).

B.) A simple random sample with n-50 provided a sample mean of 21.5 and a sample standard deviation of 4.6 .

a. Develop a 90% confidence interval for the population mean (to 1 decimal).

,

b. Develop a 95% confidence interval for the population mean (to 1 decimal).

,

c. Develop a 99% confidence interval for the population mean (to 1 decimal).

Answer 1

Answer:

1.

Given,

n = 57

xbar = 33.79

s = 14

confidence interval = 99%

Here for 99% confidence interval , z value is 2.58

a)

Margin of error = E = z*s/sqrt(n)

substitute the given values in the above formula

= 2.58*14/sqrt(57)

= 4.7842

E = 4.78

b)

Now to give the 99% confidence interval

Interval = xbar +/- E

substitute values

= 33.79 +/- 4.78

= (33.79 - 4.78 , 33.79 + 4.78)

= (29.01 , 38.57)

I hope it works for you.

Please post the second question as separate post.