# DBA-830 Module 3 Problem Set Problem 1 essay

College of Doctoral Studies

DBA-830:Module 3 Problem Set

Problem1:

InModule 2, you learned how to compute a z-score from a raw score. Inthis module, you are shown how to estimate the probability of gettinga certain z-score value equal to or higher than the one that isobserved (i.e., more extreme in the tail), as well as the proportionof all z-values that would NOT be in the tail of the distribution ofall possible z-scores in a normally shaped distribution. To do this,you compute the z-score value and then look up theprobabilities/proportions that match that z-score in Table B.1 in theback of your textbook.

(Seepp. 178-179 and pp. 699-702 in textbook.)

z-score= (X – Mean) / Std. Dev.

 Raw X Value Mean SD z-score Prop./Prob. in tail Prop./Prob. in body A. 38.25 30 5 1.65 0.0495 0.9505 B. 39.80 30 5 1.96 0.025 0.975 C. 17.00 14 1.5 2.00 0.0228 0.9772 D. 11.5 15 1.5 -1.67 0.0475 0.9525 Summary Which of the above raw score/z-values (a, b, c, and/or d) would be extreme enough to occur 5% or less of the time (i.e., p < .05) within its distribution of scores?
1. Z-score = 38.25 – 30 / 5 = 1.65

2. Z-score = 39.80 – 30 /5 = 1.96

3. Z-score = 17.00 – 14 / 1.5 = 2.00

4. Z-score = 11.5 – 15 / 1.5 = -1.67

Fromthe above raw scores, all the scores in a,b,c and d would occur at5%, this is because their probability is less than 5% and hence haveless than a 5% chance of occurrence.

Problem2:

Anactual outcome can be compared with the probability of getting thatoutcome by chance alone. This is the basis of inferential statistics.In inferential statistics, we are comparing what we really observewith what would be expected by chance alone. That which would beexpected by chance alone would be the null hypothesis (that is,nothing is going on here but chance alone).

Ifwe were to throw a coin, there would be a 50% chance it would come upheads, and a 50% chance it would come up tails by chance alone. Byextension, if we threw the coin 20 times, we’d expect 50% (p= .5 or pn = 20*.5 = 10) of the tosses to come up heads, and 50% (q= .5 or qn = 20*.5 = 10) to come up tails by chance alone if this isa fair coin.

1. You aren’t sure if your friend is using a fair coin when he offers to toss the coin to decide who will win \$100. You ask him to let you toss the coin 25 times to test it out before you decide whether you will take the bet, using this coin. You toss the coin 25 times and it comes up heads 19 times. Is this a fair coin (the null hypothesis)? What is the probability of getting 19 heads in 25 tosses by chance alone? You have decided that if the outcome of 19/25 tosses as heads would occur less than 5% of the time by chance alone, you will reject the idea that this is a fair coin. (see pp. 184-196 and 699-702 in textbook).

Pn= .5(25) = 12.5 qn= .5(25) = 12.5 hence binomial distribution can beused

Sincethey are both greater than 10.

Mean= pn = 12.5

Std.Dev. = √ npq = √25*0.5*0.5 = √62.5 = 2.5

Z-score= (X – pn)/ Std. Dev = 19 – 12.5 / 2.5 = 2.6

Fromthe table B.1, the Z-score of 2.60 is 0.0047 = 0.47%. Therefore, the

Probabilityis less than 5%. Hence the null hypothesis which would suggest that the coin is fair should be rejected.

1. Now, suppose the outcome of your trial tosses was 15 heads in 25 tosses. What is the probability of 15 heads in 25 tosses? Would you decide this is a fair coin, using the 5% criterion as in question a (see pp. 184-196 and 699-702 in textbook)?

pn= .5(25) = 12.5 qn = .5(25) = 12.5 hence binomial distributionshould be used as they are both greater than 10.

Mean= pn = .5 (25) = 12.5

Std.Dev = √npq = √25*0.5*0.5 = √ 62.5 = 2.5

Z-score= X – pn / Std. Dev = 15 – 12.5 / 2.5 = 1.00

Fromthe Table, the Z-score of 1.00 is 0.1587, which is 15.87%. Therefore,the probability is more than 5%. Hence, the null hypothesis thatwould suggest that the coin is fair should not be rejected as it ismore than 5%, the region in the tail.

Problem3:

Ateacher, Mrs. Jones, tests her 8th grade class on astandardized math test. Her class of 20 students (n) gets a meanscore (M) of 80 on the test. She wants to know how her class did incomparison with the population of all 8th grade classesthat have taken this test. She goes to a national database and findsout that the national average ()of scores for the population of all 8th graders who tookthis test is 78, with a population standard deviation (of 3 points.

1. Based on the population mean and standard deviation, what is the expected mean and standard deviation (standard error) for the distribution of sample means based on the sample size of 20 students in a class? (See pp. 201-211)

Std.error of mean = pop. Std. Dev/ √ Sample Size = 3/ √20 = 3/ 4.47 =0.67

1. If this distribution of the sample means is normal, what would be the z-score equal to a mean test score of 80 that Mrs. Jones’ class received? (See pp. 211- 215, Table B.1 in textbook)

Z-score= X – Mean / Std. Dev = 80 -78 / 0.67 = 2.98

1. When you look up the z-score you computed in part b, what is the probability of obtaining a sample mean greater than M = 80 for a sample of 20 in this population? (See Table B.1 in textbook)

Theproportion in the tail would be 0.0014 which is 0.14%. Therefore,0.14% of

themeans in the distribution would be greater than or equal to 80.

1. Mrs. Jones wants to know if her class did significantly better than the average 8th grade class on this test. (See Chapters 7 and 8)

• What is the null hypothesis?

• The mean score of Mrs. Jones class is not different from the mean score of

• All 8th graders who took the test.

• What is the alternative hypothesis?

• The mean score of Mrs. Jones class is significantly different than the mean

• Score of all 8th graders who took the test.

• Is the mean score obtained in Mrs. Jones’ class (sample) significantly different from the population mean, using the criterion that her class’s score would have to fall in the part of the distribution of all scores in the population that is above the mean and has frequencies of occurrence of 5% or less of all scores in the population (i.e., her class’s mean score would have a probability of occurring by chance alone of p &lt .05)?

• Based on the probability given, Mrs. Jones class performance would be

• Significantly different. This is because the probability of Mrs. Jones class

• Is 0.0014 which is less than 0.05 hence the null hypothesis should be

• Rejected since the 0.14% is within the tail region.