Question

3. [10 marks] A sample survey of 54 discount brokers showed that the mean price charged for a

trade of 100 shares at $50 per share was $33.77 and a sample standard deviation of $15.

a.    [3] Develop a 95% confidence interval for the mean price charged by discount brokers for a trade of 100 shares at $50 per share.

b.    [2] Explain, in context, what the interval you found tells you.

c.     [3] What sample size would be necessary to achieve a margin of error of $2? Assume a

confidence level of 95%.

d.    [2] Three years ago the mean price charged for a trade of 100 shares at $50 per share was

$39.25. Has the price dropped significantly? Justify.

Answer 1

3 a). For the given sample where population standard deviation is not given we use T-distribution so:

The confidence interval is calculated as:

μ = M ± t(s_M)

where:

M = sample mean
t = t statistic determined by the confidence level at a degree of freedom=n-1 by T table shown below
s_M = standard error = √(s²/n)

M = 33.77
t = 2.01
s_M = √(15²/54) = 2.04

μ = M ± t(s_M)
μ = 33.77 ± 2.01*2.04
μ = 33.77 ± 4.0942

95% Confidence Interval is [29.6758, 37.8642].

b), Interpretation:

We can 95% confident that the mean price charged for a trade of 100 shares at $50 per share is in the above-calculated confidence interval.

c) The minimum sample is calculated as:

Where E= margin of error and at 95% confidence interval the Z score is computed as 1.96 from the Z table shown below.

thus the minimum sample required is 216.

d) Now if the mean price charged for a trade of 100 shares at $50 per share was $39.25 and the confidence interval says it has to be in [29.6758, 37.8642] thus we can see that the interval is below $39.25 so, we can say that at 95% confidence that the price has decreased significantly.

The T and Z table as: