review-on-testing

Step2 recognize that it will be a t test with 24 df

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Unformatted text preview: faction Fund High Low Bond 15 Stock TaxDef All Med All 3 12 30 24 2 4 30 1 15 24 40 40 20 40 100 Problem 5- con8nued H0: Client satisfaction is independent of fund type Ha: Client satisfaction is not independent of fund type ˆ (f ij − Eij )2 2 χ=∑ ; ˆ Eij all cells rT c M ( 40)( 40) (30)( 40) EBH = = = 12 , ... , ETM = = = 16 n 100 n 100 (15 − 12)2 (3 − 6)2 ( 24 − 16)2 Client Satisfaction = + + ... + 12 6 16 Fund High Low Med All 15 3 12 30 = 0.7500 + 1.5000 + ... + 4.0000 = 46.4375 Bond rBc H Critical − value = 11.345 df = (3- 1)(3- 1) = 4 Stock TaxDef All REJECT NULL 12 24 12 1 16 40 6 2 6 15 8 20 12 4 12 24 16 40 30 40 100 Problem 6 An insurance company is reviewing its current policy rates. When originally setting the rates they believed that the average claim amount was $1,800. They are concerned that the true mean is actually higher than this, because they could potentially lose a lot of money. They randomly select 25 claims, and calculate a sample mean of $1,950 and standard deviation of $300. Based on a suitable hypothesis test at 5% level, do you _ind suf_icient evidence in the sample for company to be concerned? Tick the correct answer: Yes No Don’t have suf_icient data to conduct a correct statistical test H A : µ > 1800 Step 1: write the alternative hypothesis correctly and recognize that it’s a 1- sided greater than alternative. Step2: recognize that it will be a t test with 24 df. Step 3: check if conditions for the tests are met. If yes proceed with the tests else choose ‘need more data’ option. Large Sample , so we are good here. Step 4: Critical value at 5% level=1.711 (from t table using one- tail). Calculate: t= x − µ 0 1950 − 1800 = = 2.5 sn 200 25 Step5: Reject null hypothesis as 2.5 > 1.711....
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This note was uploaded on 01/15/2014 for the course BUAD 310 taught by Professor Lv during the Fall '07 term at USC.

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