Question

1) A liquor store owner wants to re-order their store to put the more expensive wines near the front. The owner samples 40 bottles of red and 40 bottles of white wine to find out which is more expensive. The owner finds the average price of white is $19.33, while the average price of red is $20.87 (standard deviation of $2.88 for white and $3.05 for red).

1a) What is the point estimate of the population price difference between red and white wines?
1b) What is the margin of error for alpha=0.01?
1c) What is the confidence interval for the difference between the two population means? Use alpha=0.05.

Answer 1

Bellow

Explanation:

1a) The point estimate of the population price difference between red and white wines is given by:

$$\bar{x}_1 - \bar{x}_2 = 20.87 - 19.33 = 1.54$$

Therefore, the point estimate of the population price difference between red and white wines is $1.54.

1b) The margin of error for alpha=0.01 can be calculated using the following formula:

$$ME = z_{\alpha/2}\cdot\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$$

where $z_{\alpha/2}$ is the critical value of the standard normal distribution for the given level of significance, $s_1$ and $s_2$ are the sample standard deviations, and $n_1$ and $n_2$ are the sample sizes for the two populations.

For alpha=0.01, the critical value of the standard normal distribution is $z_{\alpha/2}=2.58$. Substituting the given values, we get:

$$ME = 2.58\cdot\sqrt{\frac{2.88^2}{40} + \frac{3.05^2}{40}} \approx 1.01$$

Therefore, the margin of error for alpha=0.01 is approximately $1.01.

1c) The confidence interval for the difference between the two population means can be calculated using the following formula:

$$(\bar{x}_1 - \bar{x}_2) \pm z_{\alpha/2}\cdot\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$$

For alpha=0.05, the critical value of the standard normal distribution is $z_{\alpha/2}=1.96$. Substituting the given values, we get:

$$(20.87 - 19.33) \pm 1.96\cdot\sqrt{\frac{2.88^2}{40} + \frac{3.05^2}{40}}$$

Simplifying this expression, we get:

$$(1.54) \pm 1.11$$

Therefore, the 95% confidence interval for the difference between the two population means is approximately $(0.43, 2.65)$.