Question

In order to ensure efficient usage of a server, it is necessary to estimate the mean number of concurrent users. According to records, the average number of concurrent users at 100 randomly selected times is 37.7. The population standard deviation is σ = 9.2.

(a) Construct a 90% confidence interval for the expectation of the number of concurrent users.

(b) Conduct a hypothesis test to test whether the true mean number of concurrent users is greater than 35. Based on your hypothesis test, do you have evidence that the true mean number of concurrent users is greater than 35?

Answer 1

Solution

Given that

a ) = 0.10  

= 35

= 37.7

= 9.2

n = 100

This is the right tailed test .

The null and alternative hypothesis is ,

H₀ :   = 35

H_a : >  35

Test statistic = z

= ( - ) / / n

= (37.7- 35) /9.2 / 100

= 2.935

Test statistic z value = 2.935

P(z > 2.935)

= 1 - P(z < 2.935 )

= 1 - 0.9983

= 0.00017

P-value = 0.0017

= 0.10  

P-value <

0.0017 < 0.10

The null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 35, at the 0.10 significance level.