Hypothesis Testing Practice Problems(1)

scientist believes that andomly selected from IQ in sample (Xbar)= 1 reject the null Ho at th 00 00

EXCEL FORMULA Given Given Given #N/A /sqrt(n) =? Given #N/A=(Xbar )/SE -? #N/A #N/A Yes, she can reject th confidence level bec for the sample Z-sco value is less than alp Using critical Z value 2.5 is more extreme 1.64 =0.05) and 2. (? null hypothesis can b 95% and 99% confid

he Ho at the 95 and 99% cause the p-value is 0.006 ore. In both cases, the p- pha. es, the sample Z-score of e than a critical Z value of .33 =0.01). Thus, the (? be rejected at both the dence level.

6

84 920.79 85 941.75 86 937.89 87 1024.13 88 952.33 89 879.74 90 866.77 91 1080.00 92 1002.06 93 1038.74 94 1017.37 95 988.42 96 893.74 97 1022.92 98 1081.83 99 1154.07 100 827.85

Cable Breaking Strength 1 4919.00 2 5048.09 3 5482.85 4 5461.08 5 4583.24 6 4926.37 7 5460.99 8 5214.26 9 5286.32 10 4767.98 11 3931.76 12 5191.09 13 5453.64 14 4543.81 15 6060.19 16 6356.22 17 5306.83 18 4515.58 19 4713.03 20 4827.88 21 5873.37 22 4474.94 23 4659.81 24 5255.17 25 5216.45 26 5929.35 27 5258.75 28 4797.17 29 5060.31 30 4434.27 31 5359.47 32 5684.72

33 4959.64 34 4492.24 35 4407.15 36 4936.43 37 5928.43 38 5796.95 39 5470.01 40 5347.04 41 5245.09 42 4723.65 43 5077.84 44 5785.64 45 4390.91 46 5402.91 47 5101.36 48 5471.37 49 5823.75 50 6123.87 51 5168.42 52 4829.37 53 4870.35 54 4884.67 55 4822.61 56 5019.64 57 5560.57 58 5815.23 59 4922.61 60 4393.84 61 5500.39 62 4957.40 63 5258.44

64 5591.10

Employee Before Score After Score 1 84 84 2 66 85 3 84 87 4 57 84 5 67 88 6 80 86 7 71 81 8 68 85 9 73 90 10 83 78 11 62 77 12 63 77 13 74 83 14 64 88 15 84 80 16 81 80 17 54 83 18 61 88 19 81 80 20 86 88 21 75 93 22 71 78 23 86 82 24 74 84 25 65 86 26 90 89 27 72 81 28 71 90 29 85 86 30 66 92 31 59 81 32 69 82 33 49 78 34 61 83 35 62 82 36 58 92 37 87 87 38 73 80 39 66 87 40 76 85 The managing partner of a major consulting firm is trying to assess the effectiveness of expensive computer skills trainings given to all new entry-level professionals. In an effort to make such an assessment, she administers a computer skills test immediately before and after the training program for each of 40 randomly chosen employees. The pre-training and post-training scores of these 40 individuals are given on this sheet. Do the given sample data support the claim at the 10% significance level that the organization's training program is increasing the new employees working knowledge of computing? What about at the 1% significance level.

Patient Length 1 1 2 1 3 1 4 1 5 2 6 2 7 2 8 2 9 1 10 1 11 1 12 1 13 1 14 1 15 1 16 1 Defines Boundaries using alpha and d 17 1 Derived from Sample data (O-E)/SE 18 1 Derived from your t statistic 19 1 20 1 21 1 22 1 23 1 24 1 25 1 26 1 27 1 28 1 29 1 30 1 31 2 This data sheet contains the number o 2005). Before health insurance rules w number of days spent in a hospital by For a 5% level of significance, do the d time in the hospital after giving birth t for the test.

32 2 33 2 34 2 35 2 36 2 37 2 38 2 39 2 40 3 41 3 42 1 43 1 44 1

Ho:avg LOS >/2 days Hypothesized μ Ha: avg LOS <2 days alpha t critical #N/A t statistic #N/A p-value #N/A of days 44 mother spent in the hospital after giving were changed (the change was effective January 1, 2 anew mother was 2 days. data indicate the research hypothesis that women a than they were prior to 2005? Explain your answer

Length Mean 1.386 Standard Error 0.087 Median 1 Mode 1 Standard Deviation 0.579 Sample Variance 0.336 Kurtosis 0.589 Skewness 1.22 Range 2 Minimum 1 t-statistic Maximum 3 tcritical, decision making Sum 61 Count 44 df birth (in the year 2005), the average are now spending less in terms of the p-value

Used for Xbar, t-statistic SE=s/sqrt(n), t-statistic s (sample standard deviation), SE=s/sqrt (n) df=n-1 #N/A

Observed Heart Rate Assigned Gender Low High Total Female 11 7 18 Male 17 5 22 Total 28 12 40 Expected (Use MMULT formula below) Heart Rate Assigned Gender Low High Total Female 0 Male 0 Total 0 0 0 df =(2-1)*(2-1) =CHISQ.TEST(B5:C6,B12#) =CHISQ.INV.RT(B17,B16) =CHISQ.INV.RT(0.1,B16) =CHISQ.INV.RT(0.05,B16) MMULT formula* =MMULT(D5:D6,B7:C7)/D7 *Select cells B12:C:13 and in the 1st cell (B12), type the MMULT formula above How to use MMULT formula 1. In the Expected R x C table select B12:C13 (gray boxes) Heart Rate MMULT formula* Low High Total Female 0 Male 0 Expected Table Total 0 0 0 2. Without de-selecting the the cells, type the following in the formula box =MMULT(D5:D6,B7:C7)/D7 D5:D6 and B7:C7 are the row and column totals (orange box below), D7 is the Grand Total (green box) o Heart Rate Assigned Gender Low High Total Female 11 7 18 Male 17 5 22 Observed Table Total 28 12 40 X 2 test (p-value) X 2 statistic Critical X 2 ( =10%) ⍺ Critical X 2 ( =5%) ⍺

3. Press return. If done correctly the totals for the rows, columns and grand total for the EXPECTED table Heart Rate Assigned Gender Low High Total Female 12.6 5.4 18 Male 15.4 6.6 22 Expected Table completed Total 28 12 40 Source for data: https://saylordotorg.github.io/text_introductory-statistics/s15-01-chi-square-tests-for-i

7 of the OBSERVED counts Test-Statistic for Testing the In ?^2=∑▒ 〖 (𝑂−𝐸) 〗 ^2/𝐸 inclusive of all of the core cells If the two variables are indepe then 〗 〖 〗 ?^2 follows a chi-s

e should be the same as the OBSERVED table. independe.html

ndependence of Two Categorical Variables s of the table (not the totals). endent and O is at least 5 in each cell square distribution with (R-1)(C-1) df.

Observed Observed School 1 School 2 School 3 School 4 School 5 Paid 35 52 62 35 21 205 Default 32 33 42 17 13 137 67 85 104 52 34 342 Expected Expected School 1 School 2 School 3 School 4 School 5 Paid 205 Equation fo Default 137 =MMULT(R 0 0 0 0 0 342 Rows Columns df =(C14-1)*(C15-1) =CHISQ.TEST(C4:G5,C10:G11) =CHISQ.INV.RT(C17,C16) X 2 p-value X 2 statistic N Ho: Sc Ha: Scho

or Expected Row Totals, Column Totals)/ Grand Totals Null Hypothesis for X 2 test for Independence chool and Default Rate Independent ool and Default Rate not independent

Observed Values Oldest In-Between 127 123 345 209 Total 472 332 Expected Values Formula for expected values to use Oldest In-Between ### ### Total 472 332 df= (R-1)(C-1) =(R-1)*(C-1) X2 Test (p-value) =CHISQ.TEST(D3:G4,D9#) X2 Statistic =CHISQ.INV.RT(D14,D13) Expected Values Formula for expected values to use Oldest In-Between Most Delinquent Least Delinquent Most Delinquent Least Delinquent Most Delinquent Least Delinquent Null Hypothesis for X 2 test for Independence Ho: Birth Order and Delinquency Independent Ha: Birth Order and Delinquency not independent

Total 472 332

Youngest Total 93 17 360 158 65 777 251 82 ### =MMULT(H3:H4,D5:G5)/H5 Youngest Total 360 770 251 82 ### =MMULT(H3:H4,D5:G5)/H5 Youngest Total 360 770 Only Child Only Child Only Child How to calculate the ex Method 1: R*C/N. Cell D9: (472*360)/1137=149.45 Cell D10: (472*770)/1137= 319.65 How to calculate the e Method 2: Select the rows and co In the first selected cell type: =MM

251 82 ###

xpected values: expected values: olumns to be filled in MULT(H3:H4,D5:G5)/H5

Observed Diet Cancers Fatal heart diseaseNon-fatal heart diseaseHealthy AHA 15 24 25 239 Mediterranean 7 14 8 273 Total 22 38 33 512 Expected using =MMULT (Row Totals, Column Totals)/Grand Total Diet Cancers Fatal heart diseaseNon-fatal heart diseaseHealthy AHA Mediterranean Total 0 0 0 0 p-value #N/A TEST-STATISTIC #N/A Question: Is there a significant relationship between Diet and Outcome? Ho: there is no significant relationship. Ha: there is a significant relationship.

Total 303 302 605 Total 303 302 605