Question

Two file servers are compared according to their response time for retrieving a small file. The mean response time of 50 such requests submitted to server 1 was measured to be 682ms with a known standard deviation of 25ms. A similar measurement in server 2 resulted in a sample mean of 676ms with a standard deviation of 28ms. Do these samples provide sufficient evident to conclude that server 1 provides better response than server 2. Perform this test at 0.05 level of significance.

Answer 1

Answer on question #41941, Math, Statistics and Probability

Two file servers are compared according to their response time for retrieving a small file. The mean response time of 50 such requests submitted to server 1 was measured to be 682ms with a known standard deviation of 25ms. A similar measurement in server 2 resulted in a sample mean of 676ms with a standard deviation of 28ms. Do these samples provide sufficient evident to conclude that server 1 provides better response than server 2. Perform this test at 0.05 level of significance.

Solution

Step 1. State $H_{o}$ : $\mu_{1} \leq \mu_{2}$ , $H_{1}$ : $\mu_{1} > \mu_{2}$ .

Step 2. Type of test - right- tailed test.

Step 3. Level of significance: $\alpha = 0.05$

Step 4. Critical value of the statistic: $z = 1.6450$ .

Step 5. Diagram

Step 6. Decision rule: Reject $H_{o}$ if $z$ computed from evidence is more than 1.6450.

Step 7. Compute the statistic:

Evidence: $n_1 = n_2 = n = 50$ , $\bar{x}_1 = 682$ , $\bar{x}_2 = 676$ , $\sigma_1 = 25$ , $\sigma_2 = 28$ .

$$z = \frac {\bar {x} _ {1} - \bar {x} _ {2}}{\sqrt {\frac {\sigma_ {1} ^ {2} + \sigma_ {2} ^ {2}}{n}}} = \frac {6 8 2 - 6 7 6}{\sqrt {\frac {2 5 ^ {2} + 2 8 ^ {2}}{5 0}}} = 1. 1 3 0 3.$$

Step 8. Conclusion:

Do not Reject $H_{o}$ . These samples doesn't provide sufficient evident to conclude that server 1 provides better response than server 2.

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