Confidence interval

Question

A sample survey of 60 discount brokers showed the mean price charged for a trade of 100 shares at N$ 500 per share was N$ 337. The survey is conducted annually. With the historical data available, the population variance is estimated to be 2250 (N$ squared).

REQUIRED:

(i) Develop and interpret a 97% confidence interval for the mean price charged by discount brokers for a trade of 100 shares at N$ 500 per share.

(ii) Suppose that a 98% confidence interval for the mean price charged by discount brokers for a trade of 100 shares at N$ 500 per share is to be estimated with a margin of error of N$ 5. How large should the sample size be?

Answer 1

i) Using the 7:ZInterval Test under STAT > TESTS > 7 on a TI-84,

ZInterval

Inpt: Data Stats

σ:sqrt(2250) auto-completes to 47.434164902...

x-bar: 337

n: 60

C-Level: .97

Calculate

OUTPUT

ZInterval

(323.71, 350.29)

x-bar=337

n=60

ii) alpha = 0.02, so z_α/2 = -invNorm(.01,0,1) = 2.3263

n >= (z_α/2)²σ²/E² = (2.326)^2 (2250) / 5² = 487.07 and we always round up here, so n >= 488.

Hope these helped! Let me know if I can explain any of the steps in more detail. Thank you!