Question

gnt wll take 20 or less minutes to get 6. (10 marks) AbeNet, an Internet service provider 0SP), has esperieeed rapid guowth in and past five years. As a part of its marketing strategy, AbcNet promists dependable service. To achieve its obiectives. AbcNet constantl its servers. On fast connections and y evaluates the capacity of e component of its evaluation is an analysis of the average amount of time connected and using the Internet daily. A random sample of 12 customer a customer is records shows the following daily usage times, in minutes: 268 336 296 311 306 335 301 278 290 393 373 329 sing the sample data, compute the best point estimate of the population mean for daily usage times for AbcNet's customers. arks) The managers of AbcNet's marketing department would like to develop a 99% confidence interval estimate for the population mean daily customer usage time. Because the population standard deviation of daily customer usage time is unknown and the sample size is small, what assumption must the marketing managers make concerning the population of daily customer usage times? (c)(4 marks) Construct and interpret a 99% confidence interval for the mean daily usage time for AbcNet's customers.

Answer 1

(a) Sample mean = (268+336 +296 +311+306+335+301+278+290+393+373+329)/12 = 3816/12 = 318

(b) We must assume that the population follows a normal distribution to estimate the 99% confidence interval.

(c) If the population standard deviation is not known and total observations (n) are less than 30, then the sample mean follows a t-distribution with (n-1) degrees of freedoms. The formula is stated as below -

$(\bar{x} - t_{0.005,n-1} \times SE(\bar{x}), \bar{x} + t_{0.005,n-1} \times SE(\bar{x}))$

$\bar{x} = 318$ (sample mean)

is standard error of sample mean.

SE(T)

$SE(\bar{x}) = S/n^{0.5}$ where S is standard deviation of X and n is number of observation

S can be calculated as 37.21.

SE(z) = 37.21/120.5-1074

t0.005,11
is the t-critical at 1% level. Using t-table,
to. 005,11 = 3.10581

Substituting it in formula for 99% confidence interval -

$(\bar{x} - t_{0.005,n-1} \times SE(\bar{x}), \bar{x} + t_{0.005,n-1} \times SE(\bar{x}))$

(318-(3.1058 × 10.74 16). 318+ (3.1058 × 10.7416))

(318 -33.3652,31833.3652)

is 99% confidence interval