Find accurate and reliable answers to your questions on IDNLearn.com. Find reliable solutions to your questions quickly and accurately with help from our dedicated community of experts.

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)

We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for choosing IDNLearn.com for your queries. We’re committed to providing accurate answers, so visit us again soon.