A random sample of 10 gears from Supplier 1 results in x 1 = 296 and s1 = 12. An

Question

A random sample of 10 gears from Supplier 1 results in x 1 = 296 and s1 = 12. Another random sample of 25 gears from Supplier 2 results in x 2 = 315 and s2 = 18. Assume the populations are independent and normally distributed.

Is there sufficient evidence to conclude that the variance of impact strength is different for the two suppliers at the a= 0.05 level of significance?

Is there evidence to support the claim that Supplier 2 provides gears with higher mean impact strength? Use a=0.05.

Construct a one-sided upper confidence interval for mean impact strength and explain how the interval can be used to answer #2.

Explanation / Answer

Solution:

Is there sufficient evidence to conclude that the variance of impact strength is different for the two suppliers at the a= 0.05 level of significance?

Solution:

Here, we have to use F test for checking whether the variance of impact strength is different for two suppliers or not.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: There is no any significant difference in variances of impact strength for two suppliers.

Alternative hypothesis: Ha: There is a significant difference in variances of impact strength for the two suppliers.

H0: 12 = 22 versus Ha: 12 22

This is a two tailed test.

We are given level of significance = = 0.05

(Sample with larger variance considered as first)

We are given

X1bar = 315

X2bar = 296,

S1 = 18

S2 = 12

N1 = 25

N2 = 10

Test statistic formula is given as below:

F = S12 / S22

F = 18^2/12^2 = 2.25

df1 = N1 – 1 = 25 – 1 = 24

df2 = N2- 1 = 10 – 1 = 9

Upper critical value = 3.6142

P-value = 0.2065

P-value > = 0.05

So, we do not reject the null hypothesis that there is no any significant difference in variances of impact strength for two suppliers.

There is insufficient evidence to conclude that there is a significant difference in variances of impact strength for the two suppliers.

Is there evidence to support the claim that Supplier 2 provides gears with higher mean impact strength? Use a=0.05.

Here, we have to use two sample t test with equal variances.

H0: µ1 = µ2 versus Ha: µ1 < µ2

This is a one tailed test. This is a lower tailed or left sided test.

Test statistic formula is given as below:

t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]

Where Sp2 is pooled variance

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

Sp2 = [(10 – 1)*12^2 + (25 – 1)*18^2]/(10 + 25 – 2)

Sp2 = 274.9091

t = (296 – 315) / sqrt[274.9091*((1/10)+(1/25))]

t = -3.0626

Lower critical value = -1.6924

P-value = 0.0022

(Critical value and P-value is calculated by using t-table or excel)

P-value < = 0.05

So, we reject the null hypothesis that both population means are equal.

There is sufficient evidence to conclude that supplier 2 provides gears with higher mean impact strength.

Construct a one-sided upper confidence interval for mean impact strength and explain how the interval can be used to answer #2.

One sided upper confidence interval = (X1bar – X2bar) + t*standard error

Standard error = sqrt[Sp2*((1/n1)+(1/n2))]

Standard error = sqrt[274.9091*((1/10)+(1/25))]

Standard error = 6.2038

df = 10 + 25 – 2 = 35 – 2 = 33

Critical t value = 2.348338 (by using t-table or excel)

Difference = (X1bar – X2bar) = -19

One sided upper confidence interval = (X1bar – X2bar) + t*standard error

One sided upper confidence interval = (-19) + 2.348338*6.2038 = -4.43138

Test statistic value t = -3.0626 is lower than upper one sided confidence interval. So, we reject the null hypothesis.

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