Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

Two ride-sharing companies are competing for business in a large city. To help customers compare the companies, a data scientist selects a random sample of 40 completed rides from each of the two companies. From each sample, he computes the mean amount of time that passed between when the request for a pickup was placed and when the customer was picked up. Company A took, on average, 18 minutes to pick up customers with a standard deviation of 4 minutes. Company B took, on average, 12 minutes to pick up customers with a standard deviation of 10 minutes. (a) Construct and interpret a 98% confidence interval for the difference in mean response time for these two ride-sharing companies.

Sagot :

Solution :

Given :

[tex]$n_1 = n_2 = 40$[/tex]

[tex]$\overline X_1 = 18$[/tex]

[tex]$S_1 = 4$[/tex]

[tex]$\overline X_2 = 12$[/tex]

[tex]$S_2 =10$[/tex]

So we want to test :

[tex]$H_0 : \mu_1=\mu_2 $[/tex]   vs [tex]$H_1 : \mu_1 \neq \mu_2 $[/tex]

a). For 98% confidence interval :

[tex]$=\left((\overline X_1 - \overline X_2) \pm Tn_1+n_2-2, \alpha / 2 \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\right)$[/tex]

[tex]$=\left((18-12) \pm 2.375 \sqrt{\frac{4^2}{40}+\frac{10^2}{40}}\right)$[/tex]

[tex]$=(6 \pm 4.0445)$[/tex]

= (1.956, 10.045)