In order to ensure efficient usage of a server, it is necessary to estimate the

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 sample mean and sample standard deviation of number of concurrent users at 100 randomly selected times is 37.7 and 9.2, respectively. (a) Do these data provide significant evidence, at 1% significance level, that the mean number of concurrent users is greater than 35? (use p-value) (b) Construct a 99% confidence interval for the mean number of concurrent users. To assess the accuracy of a laboratory scale, a standard weight that is known to weigh 1 gram is repeatedly weighed 4 times. The resulting measurements (in grams) are: 0.95, 1.02, 1.01, 0.98. Assume that the weighings by the scale when the true weight is 1 gram are normally distributed with mean mu. (a) Do these data give evidence at 5% significance level that the scale is not accurate? Answer this question by performing an appropriate p-value test of hypothesis. (b) Use these data to compute a 95% confidence interval for mu.

Explanation / Answer

9)

a)
Hypothesis:

Ho : mu = 35
Ha : mu > 35

Test statistic :
x = 37.7 , std . deviation = 9.2 , n = 100

z = ( x - mean) / ( s / sqrt(n))
= ( 37.7 - 35) / ( 9.2 / sqrt(100))
= 2.935

we need to find p value using z = 2.935
p value = 0.0016

yes, these data provide significant evidence at 1% significance level

b) z value at 99% CI = 2.576

CI = mean + /- z * ( s /sqrt(n))
= 35 + / - 2.576 * ( 9.2 / sqrt(100))
= (32.630 , 37.369)

10)

Hypothesis:

Ho : mu = 1
Ha : mu not equal to 1

Test statistic :
x = 0.99 , std . deviation = 0.0316 , n = 4

t = ( x - mean) / ( s / sqrt(n))
= ( 0.99 - 1) / ( 0.0316 / sqrt(4))
= -0.6324

we need to find p value using t = -0.6324 , df = 3

p value = .00001.

yes, these data provide significant evidence at 5% significance level

b) t value at 95% CI = 3.182

CI = mean + /- z * ( s /sqrt(n))
= 1 + / - 3.182 * ( 0.0316/ sqrt(4))
= (0.9497, 1.0503)

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