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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 Fall '07
 Lv
 Null hypothesis, Harshad number

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