# Spring 12 QMB3200 FSU Exam 3 Practice Problems Solutions

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QMB3200  Exam  3  Practice  Problems  Solutions       1. Answer:    (E)     As  per  the  formula  sheet,  the  degrees  of  freedom  for  a  two-­‐‑sample  t-­‐‑test  are…     n1 + n2 − 2 = 15 + 18 − 2 = 31       2. Answer:    (C)     Since  we  do  not  know  the  population  standard  deviations  and  the  sample  sizes  are   small,  we  have  to  do  a  two  sample  t-­‐‑test.     For  this  question,  you  first  have  to  get  the  pooled  standard  deviation:     sp =
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## Scientific Method

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1 QMB3200 Exam 3 Practice Problems Solutions 1.   Answer: (E) As per the formula sheet, the degrees of freedom for a two-sample t-test are… + − = + − = 1 2 2 15 18 2 31 n n   2.   Answer: (C) Since we do not know the population standard deviations and the sample sizes aresmall, we have to do a two sample t-test.For this question, you first have to get the pooled standard deviation: = − + − − + − + − + − = = = 2 2 2 21 1 2 21 2 ( 1) ( 1) (15 1)2.5 (18 1)3.1250.872.842 15 18 2 31  p n s n ssn n  Then, use that in the formula for the test statistic: − −− − − =+ + + = = 0stat1 2 1 2 ( ) H value(38 45) 0 71 1 1 1 1 12.84 2.8415 18 15 18  p x xtsn n  Note: We know that the H 0 value is 0 (as it is 99% of the time) because we are simplytesting for a ‘difference’ in the scores. 3.   Answer: (C) In this scenario, the population standard deviations are unknown. However, since bothsample sizes are above 30, we can do a two sample z test.Also, since we are attempting to determine of one credit card yields a HIGHER revenue,we must be doing a one-tailed test.  2 4.   Answer: (D) Since we are doing a z test, we know that the reject H 0 region must be a z value.Also, by reading the question, we know that the following hypotheses are being tested:H 0 : µ H    < µ L   →   µ H  – µ L < 0 vs. H A : µ H  > µ L   →   µ H  – µ L > 0We know this because the question indicates that we wish to determine if the highinterest credit card yields MORE revenue.So, as per the formula sheet: invNorm (1 – α, 0, 1) = invNorm (1 – .05, 0, 1) = +1.645 5.   Answer: (C) Since the test statistic falls in the do not reject region, we can conclude H 0 i.e. µ H  – µ L < 0In other words, we believe µ H    < µ L   6.   Answer: (C) Remember we said at the review that you don’t need the difference to be zero. Youcould have some other difference tested for. In this case, since the pretzels claim to have“15 ounces more”, you must be testing for a difference of 15 ounces.Also, since we are doing a two-tailed test, we know H 0 : µ B – µ S = 15  3 7.   Answer: (B) As per the review, we know that the value for H 0 does not HAVE to be 0. That is, wecould be testing a different hypothesized value.In this instance,H 0 : µ Sony – µ Samsung > \$100 vs. H A : µ Sony – µ Samsung < \$100In addition, we are given s  p = \$25. Thus, the t test statistic is given as… − −− − =+ + = 0stat1 2 1 2 ( ) H value(310 185) 1001 1 1 12510 12  p x xtsn n = 2.34 8.   Answer: (D) Tricky question. Because the question states that the population standard deviations are  known, we will be doing a z test for independent means.Likewise, we know that we are doing a two-sided test because the question wants toknow if the mean times for the processes are equal or not.So, as per the formula sheet, we know that for a z test, we get the rejection region bydoing…invNorm( α /2, 0, 1) → invNorm(.025, 0, 1) → –1.960However, since we are dealing with a two sided test, we know that the critical z valuesare +1.960 and –1.960.That is, when you draw it the do NOT reject region is –1.960 < test statistic z < +1.96  4 9.   Answer: (E) Before we do anything else, if we want to test if the mean of population 1 exceeds themean of population 2, we must be testing…H 0 : µ 1 < µ 2 vs. H A : µ 1 > µ 2  Also, be careful that in the problem we are given the variance (NOT the standarddeviation) of each population. Thus, we compute the test statistic as… − + −− + − = = =+ − + − 2 21 1 2 21 2 ( 1) ( 1)(15 1)36 (13 1)2530.922 15 13 2  p n s n ssn n = 5.561Then, use that in the formula for the test statistic: − −− − =+ + = = 0stat1 2 1 2 ( ) H value(50 40) 0 102.1071 1 1 15.56115 13  p x xtsn n = 4.75So, we now need to look this test statistic up on the t table with df  = n 1 + n 2 – 2 = 15 + 13 –2 = 260.1 0.05 0.025 0.01 0.005 1Taildf0.2 0.1 0.05 0.02 0.012Tail: : : : : :26 1.100 1.383 1.833 2.262 2.821 ↑   t = 4.75Since the test statistic falls off the t-table, we know that the one-tailed p-value is less than.005
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