Question

In a particular year, 68% of online courses taught at a system of community colleges were taught by full-time faculty. To test if 68% also represents a particular state's percent for full-time faculty teaching the online classes, a particular community college from that state was randomly selected for comparison. In that same year, 35 of the 44 online courses at this particular community college were taught by full-time faculty. Conduct a hypothesis test at the 5% level to determine if 68% represents the state in question. Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

Answer 1

Solution:

Claim : 68% represents a particular state's percent for full-time faculty teaching the online classes.

Given :

35 of the 44 online courses at this particular community college were taught by full-time faculty.

thus sample proportion = $\hat{p}=\frac{x}{n}=\frac{35}{44}=0.7955$

level of significance = $\alpha =0.05$

Step 1) State H0 and H1:

$H_{0}:p=0.68\: \: \: \: Vs\: \: \: \: \: H_{1}:p\neq 0.68$

Step 2) Find test statistic:

$z= \frac{\hat{p}-p}{\sqrt{\frac{p\times (1-p)}{n}}}$

$z= \frac{0.7955-0.68}{\sqrt{\frac{0.68\times (1-0.68)}{44}}}$

$z= \frac{0.1155 }{\sqrt{\frac{0.68\times 0.32}{44}}}$

$z= \frac{0.1155 }{\sqrt{0.0049455}}$

$z= \frac{0.1155 }{0.0703239 }$

$z=1.64$

Step 3) Find p value:

p value = 2 X P( Z > z value)

p value = 2 X P( Z > 1.64)

p value = 2 X ( 1 - P( Z < 1.64) )

Thus to get P( Z < 1.64) , look in z table for z = 1.6 and 0.04 and find corresponding area .

.00 01 02 .03 040506.0708 .09 0.0 5000 5040 5080 .5120 5 60 5199 5239 5279 5319 .5359 5517 5557 5596 5636 .5675 5714 .5753 0.2 5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 6517 . 6879 0.5 .6915 6950 .6985 .7019 .754 7088 7123 .7157 .7190 .7224 0.6 7257 .7291 7324 .7357 789 7422 7454 7486 7517 .7549 0.7 .7580 7611 7642 7673 704 7734 7764 7794 7823 7852 0.8 .7881 .7910 .7939 .7967 .7p95 .8023 .8051 .8078 .8106 .8133 81598186 8212 8238 8264 8289 8315 .8340 8365 8389 8485 8508 8531 8554 8577 8599 .8621 8810 .8830 1.2 8849 8869 8888 8907 8 25 .8944 8962 8980 .8997 .9015 1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177 1.4 .9192 .9207 .9222 .9236 .9 1 .9265 .9279 .9292 .9306 .9319 1.5 .9332 9345 9357 9370 9382 9394 9406 9418 .9429 .9441 1.694529463 .9474 94949495 .9505 .9515 .9525 .9535 .9545 9 0.1 5398 5438 5478 0.3 0.4 6179 6217 .6255 .6293 .6331 .6368 6406 .6443 .6480 . 6554 6591 6628 6664 .6100 6736 6772 6808 6844 0.9 1.0 8413 8438 8461 1.1 8643 8665 8686 8708 829 8749 .8770 .8790 8 2 2

P( Z < 1.64) = 0.9495

Thus p value = 2 X ( 1 - P( Z < 1.64) )

p value = 2 X ( 1 - 0.9495 )

p value = 2 X 0.0505

p value = 0.1010

Step 4) Decision rule : Reject H0 , if p value < 0.05 level of significance , otherwise , we fail to reject H0.

Since p value = 0.1010 > 0.05 , we fail to reject H0.

p value 0.1010 3.00 2.001.64 00 1.64 2.00 3.00 1.00

Step 5) Since we failed to reject H0 , we conclude that : 68% represents a particular state's percent for full-time faculty teaching the online classes.