At the end of 1990,1991,and 1992 the average prices of a share of stock in a money market portfolio were $34.83,$34.65 and $31.26 respectively.To investigate the average share price at the end of 1993,a random sample of 30 stocks was drawn and their closing prices on the last trading day of 1993 were observed with a mean of 33.583 and a standard deviation of 19.149.Assuming that stock prices are normally distributed,a 90% confidence interval would be _____.
A)[27.643,39.523]
B)[26.732,40.434]
C)[32.514,34.651]
D)[32.533,34.633]
E)[32.269,34.897]

Question

Quizplus · Accepted Answer

The 90% confidence interval for the mean can be calculated using the formula: $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $t_{\alpha/2}$ is the t-score corresponding to the desired confidence level (90% in this case), $s$ is the sample standard deviation, and $n$ is the sample size. Given $\bar{x} = 33.583$, $s = 19.149$, and $n = 30$, and knowing that the t-score for a 90% confidence interval with 29 degrees of freedom (n-1) is approximately 1.699, the confidence interval is calculated as follows: $33.583 \pm 1.699 \cdot \frac{19.149}{\sqrt{30}}$, resulting in the interval [27.643, 39.523].

At the End of 1990,1991,and 1992 the Average Prices of a Share