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 he 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 10 days. He finds that Route A takes an average of 49 minutes, with standard deviation 2 minutes, and Route B takes an average of 50 minutes, with standard deviation 4 minutes. Histograms of travel times for the routes are roughly symmetric and show no outliers. (alpha =0.05)
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 -3.771808 to 1.771808

Step-by-step explanation:

Here we have the confidence interval is given by the following equation;

$\left (\bar{x}_1-\bar{x}_{2} \right ) - z_{c}\sqrt{\frac{\sigma _{1}^{2}}{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2} \right ) + z_{c}\sqrt{\frac{\sigma _{1}^{2}}{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}$

$\bar x_1$ = 49 Minutes

σ₁ = 2 Minutes

$\bar x_2$ = 50 Minutes

σ₂ = 4 Minutes

n₁ = n₂ = 10 days

α = 0.05

At 95%, from the z score table , z$_c$ = $\pm$1.959964 ≈ ± 1.96

Plugging in the values Solving we get

Δμ Min = -3.771808 and Δμ Max = 1.771808.