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Question 4 Incorrect answer. Incorrect. Try again. Two suppliers manufacture a plastic gear used in a laser printer. The impact strength of these gears measured in foot-pounds is an important characteristic. A random sample of 10 gears from supplier 1 results in x Overscript bar EndScripts Subscript 1 Baseline equals 289.30 and and another random sample of 16 gears from the second supplier results in x Overscript bar EndScripts Subscript 2 Baseline equals 321.50 and Is there sufficient evidence to conclude that the variance of impact strength is different for the two suppliers

Sagot :

Akiran007idn

Answer:

Since the calculated t = -3.9886 falls in the critical region t ∝/2 ≥±2.101 we reject the null hypothesis and conclude that variance of impact strength is different for the two suppliers

Step-by-step explanation:

The given data is

Sample1 mean = x1` = 289.3

Sample 1 size = n1= 10

Sample 1 std. dev= s1= 22.5

Sample2 mean = x2` = 321.5

Sample 2 size = n2= 16

Sample 2 std. dev= s2= 21

1)Formulate the null and alternate hypothesis as

H0: σ₁=σ₂ against the claim Ha: σ₁ ≠σ₂

2)The test statistic is

t= (x1`- x2`) / √ s1²/n1+ s2²/n2

where degrees of freedom = d.f=

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

3) the significance level chosen is ∝= 0.05

4) Calculating the d.f

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

    = [21²/10 + 22.5²/16] / [(21/²10)² /9 + (22.5²/16)²/15]

= 18

5) the critical region for ∝= 0.05 for 18 d.f =  ±2.101 for two sided test

6) t= (x1`- x2`) / √ s1²/n1+ s2²/n2

t= 289.3- 321.5/√21²/10 + 22.5²/16

t= -32.2/√44.1+31.64

t= -32.2/8.0729

t= -3.9886

7) Conclusion

Since the calculated t = -3.9886 falls in the critical region t ∝/2 ≥±2.101 we reject the null hypothesis and conclude that variance of impact strength is different for the two suppliers

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