Question

Do flexible schedules reduce the demand for​ resources? A government health department experiments with a flexible​ four-day work week. For a​ year, the department recorded the mileage driven by 11 field workers on an ordinary​ five-day work week.​ Then, it changed to a flexible​ four-day work week and recorded mileage for another year. The resulting sample had a standard deviation of 1111.3. How big should the sample have been to narrow the width of the reported​ 95% confidence interval to 600 miles? Use the sample standard deviation as your standard deviation guess.

Answer 1

Solution:

Given:

Standard deviation = s =  1111.3

c = confidence level = 95%

Width of confidence interval = 600 miles

then

Margin of Error = E= Width / 2 = 600 / 2= 300 miles

We have to find sample size n.

Formula:

$n= \left ( \frac{Z_{c}\times \sigma }{E} \right )^{2}$

$\sigma$ is unknown , but we use its estimate sample standard deviation s.

We need to find z_c value for c=95% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

0.1 V123 .00 .01 .02 .03 .04 .05 .06 0.0 ,5000 15040 5080 5120 .5160 5199 39 1.5398 .5438 5478 1.5517 .5557 .5596 .636 0.2 15793 5832 ,5871 1.5910 .5948 1.5987 ,6026 0.3 .6179 .6217 .6255 .6293 .6331 1.6368 | 6406 0.4 .6554 ,6591 ,6628 .6664 .6700 .67366 16772 0.5 ,6915 ,6950 .6985 ,7019 .7054 7088 0.6 72577291 .7324 .7357 7389 7422 F454 0. 7 7580 7611 ,7642 .7673 704 7734 7764 0.8 1.788179107939 .7967 7995 ,8023 4051 09819816,8212 ,82388 .82648289 1.4315 18413 18438 ,8461 ,8485 .8508 18531 .4554 .8643 18665 8686 .8708 ,8729 ,8749 70 18849 1.8869 .888 ,8907 .8925 .894 962 1.3 19032 9049 ,9066 19082 19099 9115 1931 1.4 19192 .92079222 .9236 .9251 ,9265979 1.5 ,9332 19345 .9357 .9370 9382 19394 9606 1.6 19452 19463 19474 .9484 .9495 19505 15 1.7 ,9554 19564 19573 9582 19591 .9599 9508 1.8 19641 .9649 .9656 .9664 .9671 .96789 586 | 19 7 47- 0710 -07- 19750 2.0 ,9772 19778 19783 19788 19793 19798 19803 1.0 1.1 1.2

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z_c = 1.96

Thus

n=(2x) n = (Zc X 5 Ε

(1.96 x 1111.3 300 )

n = (7.26049)

n = 52.71476

n = 53