Answer to Question #199149 in Statistics and Probability for mandona

Question

Suppose that the mean for the current gas prices in $3.00 with the standard deviation $0.25. Construct an interval that will contain 90% of all sample means of all random samples of size 40. 2. Suppose that you asked 40 people in different parts of San Diego and found that the mean price of gas that they bought this week was $3.04. Assume that population standard deviation is known to be $0.25. Construct a 90% confidence interval for the true mean of the gas price.

Answer 1

1.

$\mu=3 \\ \sigma = 0.25 \\ n=40$

Z-value for 90% CI = 1.64

Confidence interval:

Upper limit $= \mu + Z \times \frac{s}{\sqrt{n}}$

$= 3 + 1.64\times \frac{0.25}{\sqrt{40}} \\ = 3 + 0.06 \\ = 3.06$

Lower limit $= \mu - Z \times \frac{s}{\sqrt{n}}$

$= 3 - 1.64\times \frac{0.25}{\sqrt{40}} \\ = 3 - 0.06 \\ = 2.94 \\ CI: 2.94 ≤ \mu ≤3.06$

2.

$n=40 \\ \mu = 3.04 \\ \sigma=0.25$

Z-value for 90% CI = 1.64

Confidence interval:

Upper limit $= \mu + Z \times \frac{s}{\sqrt{n}}$

$= 3.04 + 1.64\times \frac{0.25}{\sqrt{40}} \\ = 3.04 + 0.06 \\ = 3.10$

Lower limit $= \mu - Z \times \frac{s}{\sqrt{n}}$

$= 3.04 - 1.64\times \frac{0.25}{\sqrt{40}} \\ = 3.04 - 0.06 \\ = 2.98 \\ CI: 2.98 ≤ \mu ≤3.10$