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Accounting Institute Seminar *

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ACCT 3311

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Business

Date

Nov 24, 2024

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xlsx

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35

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Strong complaints Significant complaints Minor complaints North 8 133 310 South 1 50 208 Midlands 11 293 703 East 2 72 241 West 8 161 311 Total 30 709 1773 0.187588152327221 0.174844895657078 0.17315343672084 0.187588152327221 0.070521861777151 0.413258110014104 0.101551480959097 0.227080394922426 Expected Value Strong complaints Significant complaints Minor complaints North 5.2 123.0 307.5 South 2.8 65.7 164.2 Midlands 11.9 281.6 704.1 East 4.0 93.7 234.2 West 6.1 145.2 363.0 Total 30.0 709.0 1773.0 AFTER Strong/significant complaints Minor complaints No complaints North 128.2 307.5 9807.4 South 68.4 164.2 5237.4 Midlands 293.5 704.1 22457.4 East 97.6 234.2 7470.2 West 151.3 363.0 11578.7 Total 60.0 1418.0 3546.0 Strong/significant complaints Minor complaints No complaints North 141 310 9792 South 51 208 5211 Midlands 304 703 22448 East 74 241 7487 West 169 311 11613 Total 739 1773 56551 (O-E) Strong/significant complaints Minor complaints No complaints North 12.8 2.5 -15.4 South -17.4 43.8 -26.4 Midlands 10.5 -1.1 -9.4 East -23.6 6.8 16.8
West 17.7 -52.0 34.3 (O-E)^2 Strong/significant complaints Minor complaints No complaints North 164.8 6.3 235.8 South 304.2 1918.2 694.6 Midlands 110.9 1.2 89.1 East 557.9 46.1 283.1 West 313.0 2705.8 1178.3 (O/E) Strong/significant complaints Minor complaints No complaints North 1.29 0.02 0.02 South 4.44 11.68 0.13 Midlands 0.38 0.00 0.00 East 5.71 0.20 0.04 West 2.07 7.45 0.10 df 8 p value = 0.05 CHI^2 Critical 15.5073130558655 P value Calculated 33.54 we reject the null hypothesis because th
No complaints Total customers 9792 10243 5211 5470 22448 23455 7487 7802 11613 12093 56551 59063 P value 0.0003677287433 No complaints Total customers 9807.4 10243.0 5237.4 5470.0 22457.4 23455.0 7470.2 7802.0 11578.7 12093.0 56551.0 59063.0 To calculate the critical value, we use either th Total customers 10243.0 If p-value <= α 5470.0 If p-value > α, 23455.0 exp 7802.0 Strong/significant complaints Minor complaints 12093.0 North 128.161065303151 307.482501735435 113102.0 South 68.4409867429694 164.202800399573 Midlands 293.470446811032 704.090801347713 Total customers East 97.6191185683084 234.206626822207 10243 West 151.308382574539 363.017269695071 5470 Total 60 1418 23455 obs 7802 North Strong/significant complaints Minor complaints 12093 South 141 310 59063 Midlands 51 208 East 304 703 West 74 241 Total 169 311 739 1773 nd the chi-square p-value with the help of the following formula.
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0.000367728743 he alpha α (0.05) is bigger than 0.000367729 so we accept the alternative hypothesis
he chi-square critical value table or the CHISQ formula.CHISQ.INV.RT or α, the null hypothesis is rejected. , the null hypothesis is accepted. pected No complaints Total customers 9807.35643296142 10243 5237.35621285746 5470 22457.4387518413 23455 7470.17425460949 7802 11578.6743477304 12093 3546 113102 bserved No complaints Total customers 9792 10243 5211 5470 22448 23455 7487 7802 11613 12093 56551 59063 “=CHITEST(actual_range,expected_range)”
720 720 locking the column 720 locking the row expected data E observed data O chi 2 formula ∑(O-E)^2/E df formula (row -1)*(column -1) (5-1)*(3-1) 4*2 = 8
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day quality 7 1 6 1 observed 6 2 Count of quality Column Labels 6 3 Row Labels 1 2 5 1 1 22 8 6 1 2 30 14 7 1 3 38 18 3 1 4 62 20 6 1 5 75 26 6 1 6 98 38 6 2 7 45 14 7 1 Grand Total 370 138 7 1 6 1 370 138 6 1 expected/ expected range 1 2 1 2 1 24.625623960067 9.1846921797005 1 1 2 33.24459234609 12.3993344425957 6 1 3 43.094841930117 16.0732113144759 5 1 4 58.485856905158 21.8136439267887 5 1 5 70.798668885191 26.4059900166389 5 3 6 99.118136439268 36.9683860232945 7 1 7 40.63227953411 15.1547420965058 3 3 2 1 0.735115476041235 p value 7 1 1 3 6.89390117967558 4 1 observed/ actual range 4 1 1 22 8 7 1 2 30 14 7 1 3 38 18 2 1 4 62 20 5 1 5 75 26 6 2 6 98 38 5 1 7 45 14 3 1 Grand Total 370 138 4 2 3 1 7 3 1 1 4 1 6 2 2 2 1 3 7 2 5 1 6 2
5 1 5 2 6 3 5 2 7 1 3 2 2 2 6 1 2 3 1 1 6 2 4 1 1 1 6 1 1 1 6 1 5 2 1 2 7 1 7 1 5 1 4 1 7 1 4 1 4 3 1 1 5 2 5 3 6 2 5 1 7 1 4 1 3 1 5 1 6 1 4 1 4 1 5 1 3 1 1 1 6 2 5 2 6 3 4 3 2 2 6 1 3 2
4 2 5 2 4 1 5 1 6 2 7 2 7 1 6 3 5 3 3 1 5 1 7 1 2 2 3 1 4 1 3 3 6 1 6 3 4 3 5 1 5 2 4 1 1 1 6 2 5 3 6 2 5 1 6 1 6 1 5 1 4 1 4 2 5 2 5 3 4 2 2 1 1 2 5 1 3 1 5 1 6 3 6 1 6 2 4 2 5 1 3 2 6 2
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6 1 4 1 6 1 5 2 6 1 6 3 6 1 3 1 6 1 1 1 7 3 5 1 7 1 3 3 4 1 3 2 6 3 6 3 1 1 7 1 4 1 5 1 6 2 4 1 6 1 6 1 5 2 3 1 4 3 3 3 3 2 3 1 6 1 6 1 4 3 6 1 5 1 5 1 2 1 6 3 6 1 5 2 6 1 7 3 5 1 6 1 3 1
4 1 6 2 6 3 6 1 2 3 3 1 3 1 5 3 4 2 5 1 4 2 7 1 7 1 1 2 4 2 4 1 7 2 3 2 1 3 6 2 2 1 7 2 4 3 4 1 3 2 4 1 5 1 4 1 4 1 2 2 4 1 3 2 6 1 5 2 6 1 7 2 4 3 6 3 6 3 6 3 4 2 5 1 2 3 3 1 4 1 6 1 4 3
3 2 4 3 6 1 3 1 7 1 6 1 4 1 7 1 7 1 1 2 3 3 1 1 5 3 2 1 1 1 7 1 7 3 1 1 6 3 5 1 4 1 6 1 7 3 6 1 4 1 5 1 6 1 5 2 5 1 5 1 2 1 6 1 4 2 4 1 4 1 2 1 1 3 7 1 5 1 2 1 5 2 1 3 6 1 2 1 6 1 5 2 5 1
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6 1 5 1 6 2 6 2 6 1 2 3 5 1 6 3 4 1 7 2 6 1 5 1 6 1 2 1 6 3 4 1 1 2 3 3 2 2 2 1 5 1 6 2 6 2 6 2 3 1 7 3 6 1 7 1 6 1 6 1 3 3 4 1 3 1 7 1 7 1 6 1 6 2 5 2 3 1 4 1 1 1 7 3 5 2 6 2 6 2 2 1 7 2
6 1 6 1 3 1 4 2 5 2 6 1 6 1 5 1 4 1 6 1 5 1 6 1 7 1 6 1 5 3 4 1 2 2 3 2 6 1 6 1 5 1 1 3 4 2 5 1 7 2 6 2 3 1 3 3 6 1 5 1 4 3 3 2 5 1 3 1 7 2 7 1 6 1 4 2 2 2 4 1 5 1 2 1 2 1 6 1 2 1 5 1 7 2
2 3 6 1 1 1 1 3 6 1 2 2 6 1 5 1 6 1 3 1 5 2 7 1 6 1 6 1 6 1 3 3 2 2 3 1 4 2 1 3 5 2 2 1 4 1 3 2 2 1 3 1 2 1 4 2 6 3 5 1 5 2 6 1 6 1 6 1 3 1 5 1 6 2 5 1 4 1 6 1 3 1 5 2 6 2 7 1 5 3 3 2 6 1
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5 1 5 2 7 1 4 1 2 1 6 1 4 1 5 3 4 1 6 1 3 2 1 2 7 1 4 3 7 2 7 1 2 3 6 1 1 2 5 1 6 1 4 1 6 1 3 3 6 3 2 1 3 3 3 1 4 1 2 3 5 1 5 1 5 1 4 1 3 1 6 1 5 1 6 2 3 1 5 1 7 1 7 1 3 3 1 3 7 1 5 3 1 1
4 1 2 1 3 2 5 2 7 2 5 1 7 1 5 1 2 3 2 1 2 2 3 3 6 2 6 1 2 1 5 1 2 1 7 1 4 1 7 2 7 1 6 1 4 1 6 1 5 1 3 3 4 1 4 2 1 3 7 1 6 2 6 2 6 1 2 1 6 1 3 1 6 1 6 2 6 1 4 1 5 1 4 1 3 2 6 2 1 1 6 3 3 1
1 1 1 1 6 2 6 3 6 2 2 2 5 3 5 3 3 2 2 1 6 3 5 1 6 2 4 1 6 1 2 3 5 1 3 1 5 1 5 1 4 1 2 1 5 2 4 1 6 1 5 1 3 1 4 3 5 1 7 1 6 1 7 2 3 1 5 3 5 1 6 3 4 2 6 1 6 1 5 2 2 3 4 1 7 1 3 1 4 3 4 1 6 1
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2 1 7 1 4 1 6 1 5 1 5 1 4 1 5 1 1 1 3 1 5 1 3 1 1 1 6 2 6 3 2 2 6 1 7 1 4 1 3 2 4 2 4 1 6 1 6 1 4 1 4 1 4 2 6 2 5 1 2 2 6 3 5 1 2 1 6 1 6 1 4 2 1 1 5 1
to help you go to the Canvas and Download " CL3 Excel Pivot Tables and Contingency Tables 3 Grand Total 2. Select Insert|PivotTable. This 10 40 40 Choose the “Ex 10 54 54 3. In the PivotTable Filed List 14 70 70 4. Choose “day” aga 13 95 95 “Sum of day”, choose Va 14 115 115 25 161 161 7 66 66 93 601 10. Calculate the proportions of re 11. Use these proportions to calculate the expect 93 601 12. Calculate the contributions to the chi-squar 3 6.18968386023295 8.35607321131448 10.8319467554077 14.7004991680532 17.7953410981697 24.9134775374376 10.2129783693844 6.1896839 10 40 10 54 14 70 13 95 14 115 25 161 7 66 93 601
(3).pdf" to help you go to the Canvas and Download " CL3 Excel Pivot Tables and Contingency Tables (3).pdf" 1. Click on a cell in your data set, e.g. A2 will automatically suggest using the full data set,. Including the headers. xisting Worksheet” option, specifying D2 as the location t window, click on “day” and drag it into the Row Labels box. Place “quality” in the Column Labels box. ain from the top box and drag it to the Values box. Click on the arrow to the left of lue Field Settings and select Count (you could choose “quality” and get the same result) “contingency table” that tabulates the data by day of week and quality. A chi-squared test is needed. Performing a chi-squared test? esponses for each day of the week in I4:I10, and the proportions of each quality score in E12:G12. ted frequencies in E16:H22. Check that the various sub-totals agree with the observed frequencies in the continge red statistic for each cell in E27:G33= (observed-expected)2/expected and the total chi-squared statistic in G34
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" ency table.
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Scores test whether the sample mean is significantly different from a known p 3 Calculate Sample Statistics: Calculate the sample mean and standard d 5 Sample Mean ( ): =AVERAGE(A1:A25) x̄ 4 Sample Standard Deviation (s): =STDEV.S(A1:A25) 5 Determine the population mean (μ0) based on the null hypothesis. 9 Choose a significance level (alpha) for your test, e.g., α = 0.05. 4 Z test 1.65875863875168E-09 7 9 Step 5: Interpret the Results 3 If the p-value returned by the Z.TEST function is less than your chosen 8 Remember to adjust the cell references, sample size, and significance 9 4 z test = 1.65876E-09 7 Standard Deviation = 2.43486579272276 9 0.0000000234 = 2.34E-08 3 E- means very small no. 8 9 4 7 9 3
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population mean (H0): deviation using Excel functions. In an empty cell, use the following formulas: 3 2.43486579272276 1.65876E-09 = 0.00000000165876 n significance level (α), you can reject the null hypothesis in favor of the alternative hypothesis. Thi level to match your specific data and hypothesis test. The Z.TEST function calculates the Z-statistic
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is indicates that there is a statistically significant difference between the sample mean and the population c and returns a p-value, allowing you to determine whether the null hypothesis can be rejected.
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mean.
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Student 1 Student 2 93 79 68 66 82 73 51 72 87 80 82 74 72 95 71 93 92 99 73 97 146.49 131.16 Variance Variance = 146.49 Variance = 131.16 we reject the null hy Step 3: For the “Va
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So, from the Z.TEST result, below are the results. ypothesis because the alpha α (0.05) is bigger than 0.000367729 so we accept the alter Step 1: First, we need to calculate the variables for these two values using the VAR.P function. Step 2: Go to the “Data” tab and click “Data Analysis.” Scroll down and select z-Test: Two Sample for Means” and click on “OK.” Variable 1 Range,” select “Student 1” scores. For the “Variable 2 Range,” select We can reject the null hypothesis if Z < – Z Critical Two-Tailed Z > Z Critical Two Tail . Z < – Z Critical Two Tail = -1.080775083 > – 1.959963985 Z > Z Critical Two Tail = -1.080775083 < 1.959963985
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z-Test: Two Sample for Means Student 1 Student 2 Mean 77.1 82.8 Known Variance 146.49 131.16 Observations 10 10 Hypothesized Mean Difference 0 z -1.081748 P(Z<=z) one-tail 0.139682 z Critical one-tail 1.644854 P(Z<=z) two-tail 0.279365 z Critical two-tail 1.959964 rnative hypothesis t “Student 2” score
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Detail of expenses in India(in INR) Detail of expenses in US (in INR) sample t-test using 1 t 18 19 13 10 t-test = 0.17764 11 12 16 17 15 12 17 18 16 15 15 16 18 19 12 11 17 18 14 15 13 11 10 11 14 15 15 16 13 14 14 15 16 17 12 13 10 11 TTest 0.177639610902702 Paired Samp df = (n - 1) Example 2 t-test using 1 tail distribution New Old 15 18 t-test = 0.454692 13 10 10 12 15 12 15 12 17 18 16 15 15 16 18 19 12 11 17 18 14 11 10 11 10 11 14 15 15 15 13 14 14 15 16 15 12 10 10 11
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TTest 0.454691995977266 Example 3 t-test using 1 tail distribution Control Drug t-test = 0.364818 18 19 13 10 11 12 16 17 15 12 17 18 16 15 15 16 18 19 12 11 17 18 14 15 13 11 10 11 14 15 15 16 13 14 14 15 16 17 12 13 10 11 TTest 0.364848283553589
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tail distribution Where:"n" is the number of pairs or observations in the dataset.
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