Question

A man who moves to a new city sees that there are two routes he could take to work. A neighbor who has lived there a long time tells him Route A will average 5 minutes faster than Route B. The man decides to experiment. Each day he flips a coin to determine which way to go, driving each route 20 days. He finds that Route A takes an average of 40 minutes with standard deviation 3 minutes, and Route B takes an average of 43 minutes with standard deviation 2 minutes. Histograms of travel times for the routes are roughly symmetric and show no outliers. Find a 95% confidence interval for the difference between the Route B and Route A commuting times.

Answer 1

Answer:

The 95% confidence interval for the difference between the Route B and Route A commuting times is -4.640279 < μ₁ - μ₂ <-1.359721

Step-by-step explanation:

Here we have the formula for the confidence interval of the difference between two means given as follows;

$\left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2}\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}$

Where:

$\bar{x}_{1}$ = Mean of Route A = 40

$\bar{x}_{2}$ = Mean of Route B = 43

s₁ = Standard deviation of time to work for Route A = 3 minutes

s₂ = Standard deviation of time to work for Route B = 2 minutes

n₁ = Number of days taken through Route A = 20 days

n₂ = Number of days taken through Route B = 20 days

At 95% confidence level, and df = 20 - 1 = 19 we have;

$t_{\alpha /2}$ = ±2.03452

Plugging in the values, we have;

$\left (40-43 \right )\pm 2.03452 \times \sqrt{\frac{3^{2}}{20}+\frac{2^{2}}{20}}$, which gives the 95% confidence interval for the difference between the Route B and Route A commuting times as follows;

-4.640279 < μ₁ - μ₂ <-1.359721.