Question

The distribution of times that a company's service technicians take to respond to trouble calls is normal with mean μ and standard deviation 0.25 hours. The company advertises that its service technicians take an average of 2 hours to respond to trouble calls from customers. We suspect that the average time is really more than 2 hours. From a random sample of 25 trouble calls, the average time service technicians took to respond was 2.10 hours. How strong is the evidence against the company's claim? Based on these data, the P -value of the appropriate test is (express answer with 4 places after decimal) Answer:

Answer 1

Solution:

Given:

Mean =$\mu = 2$

Standard Deviation = $\sigma= 0.25$

Sample size = n = 25

Sample mean = $\bar{x}= 2.10$

We have to test if the average service time is more than 2 hours.

Thus hypothesis are:
$H_{0}: \mu = 2$

Vs

$H_{1}: \mu > 2$

We have to find p-value:

p-value = P( Z> z test statistic )

where

o/n

$z=\frac{ 2.10-2 }{0.25 /\sqrt{25}}$

$z=\frac{ 0.10 }{0.25 / 5}$

$z=\frac{ 0.10 }{0.05 }$

Thus

p-value = P( Z> 2.00)

p-value = 1 - P( Z < 2.00)

Look in z table for z = 2.0 and 0.00 and find corresponding area.

2 .00 .01 10.0 .50 po .5040 10.1 5398 5438 10.2 .5723 5832 10.3 6179 6217 .64 .6591 6915 .6950 77 7291 1.7480 7611 7881 7910 8159 8186 8413 8438 8443 .8665 1.8869 91 32 9049 1992 9207 115 | 932 9345 11.6 9452 119463 9463 11.7 .95574 9564 11.8 9641 9649 11.9 9713 9719 12.0 - 9772 9778 .02 5080 5478 .5871 6255 .6628 .6985 7324 7642 7939 8212 .8461 .8686 8888 9066 .9222 9357 .9474 9474 9573 .9656 .9726 9783 .03 .04 .05 5120 5160 .5199 5517 5557 .5596 5910 5948 .5987 6293 6331 .6368 6664 .6700 .6736 7019 7054 .7088 7357 7389 7422 7673 7704 7734 7967 7995 .8023 8238 8264 | .8289 8485 | | 8508 .8531 .8708 1.8729 18749 8907 1.8925 18944 90821 | 9099 | 9115 .9236 9251 | 9265 9370 9382 9394 9484 9484 9495 9495 9505 .9582 9591 .9599 | .9664 .9671 | 19678 9732 9738 .9744 9788 9793 9798 .06 5239 5636 .6026 6406 .6772 7123 7454 7764 8051 .8315 8554 .8770 | 8962 9131 .9279 9406 9515 9608 9686 9750 9803 .07 .08 5279 .5319 .5675 .5714 6064 .6103 6443 6480 .6808 6844 7157 7190 7486 7517 7794 7823 8078 8106 8340 8365 8577 18599 8790 1.8810 8980 18997 9147 | 9162 9292 | 9306 9418 9429 .9525 19535 .9616 | 9625 9693 .9699 9756 9761 9808 9812 .09 | .5359 .5753 6141 6517 .6879 | 7224 1 7549 17852 | .8133 | 8389 86211 8830 9015 9177 .9319 9441 95451 | 9633 19706 1 97677 98177 |

P( Z< 2.00) = 0.9772

thus

p-value = 1 - P( Z < 2.00)

p-value = 1 - 0.9772

p-value = 0.0228