Question

The mean SAT score in mathematics, is 524. The standard deviation of these scores is 48. A special preparation course daims that its graduates will score higher, on average, than the mean score 524. A random sample of 37 students completed the course, and their mean SAT score in mathematics was 534 Assume that the population is normally distributed. At the 0.05 level of significance, can we conclude that the preparation course does what it claims? Assume that the standard deviation of the scores of course graduates is also 48 Perform a poetailedtest. Then fill in the table below Carry your intermediate computations to at least three decimal places, and round your responses as speared in the table. (if necessary, consulta list of T U V W tại P

Answer 1

Solution:

Given:

Population mean =

u = 524

Population standard deviation = $\sigma = 48$

Sample size = n = 37

Sample mean = $\bar{x} = 534$

Level of significance = 0.05

Claim: Graduates of a special preparation course will score higher than mean score 524.

Part i) State null hypothesis:

$H_{0}: \mu = 524$

Part ii) State alternative hypothesis:

$H_{1}: \mu > 524$

Part iii) Type of test statistic:

Since population is Normally distributed with known standard deviation, we use z test statistic.

Part iv) Test statistic value:

$z=\frac{\bar{x}-\mu }{\sigma /\sqrt{n}}$

$z=\frac{534-524 }{ 48 /\sqrt{37}}$

$z=\frac{10 }{ 48 /6.082763 }$

$z=\frac{10 }{ 7.891151 }$

Part v) Critical value:

Since this is right tailed test, find an area = 1 - 0.05 = 0.95.

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

.00 101 .02 103 106 107 108 209 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 13 1.4 1.5 1.6 1.7 .5000 1.5398 .5793 .6179 .6554 .6915 7257 47580 7881 18159 .413 18643 18849 19032 19192 19332 CACO 19554 15040 .5438 1.5832 .6217 .6591 .6950 7291 7611 7910 1.8186 .8438 .8665 .8869 19049 19207 19345 0463 19564 15080 15478 .5871 .6255 .6628 .6985 .7324 7642 7939 1.8212 1.8461 1.8686 18888 1.9066 1.9222 19357 19474 1973 1.5120 .5517 .5910 .6293 .6664 17019 7357 .7673 .7967 18238 .8485 8708 18907 1.9082 1.9236 19370 9484 19582 1.04 105 sl60 5199 1.5557 5996 5948 1.5987 .6331 6368 .600 .6766 7054 7088 17889 7412 .T04 734 7095 .80% .8264 18288 .8508 .851 .8/29 18749 .8925 1.894 1.9099 .9115 1951 1.925 19382 1934 19495 19505 19591 19599 15239 1.5636 .6026 .6406 1.6772 7123 7454 7764 .8051 .8315 18554 .8770 .8962 19131 1.9279 19406 19515 19608 1.5279 1.5675 .6064 .6443 1.6808 7157 7486 7794 .8078 1.8340 .8577 1.8790 1.8980 19147 1.9292 19418 9525 19616 1.5319 5714 1.6103 .6480 .6844 1.7190 7517 47823 18106 .8365 1.8599 1.810 .8997 .9162 19306 19429 19535 19625 15359 5753 .6141 .6517 .6879 .7224 7549 7852 18133 .8389 ,8621 1.8830 19015 19177 19319 1941 19545 19633

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Z_critical = 1.645

Part vi) Conclusion:

Decision Rule:
Reject null hypothesis ,if z  test statistic value > z critical value = 1.645, otherwise we fail to reject H0.

Since z  test statistic value = 1.27 < z critical value = 1.645, we fail to reject H0.

Thus at 0.05 level of significance, we can not support the preparation course's claim that its graduates score higher in SAT.

Thus answer is: No.