2. Two suppliers A and B, supplied a construction site with steel of equal diame

ID: 3055391 • Letter: 2

Question

2. Two suppliers A and B, supplied a construction site with steel of equal diameter and strength Tests were carried out by the engineer to determine the yield strength of each batch of steel. Random sampling was used in the selection process. The data obtained is presented below: Yield Strength (MPa) 41.05 33.18 36.5740.2043.09 4041 40.96 33.6447.78 3545 39.75 31.98 42.643643 3705 39.21 4149 32.1 a. Determine the 98% confidence interval on the difference in the population mean. b. Repeat question a assuming that the population standard deviation of the yield strength of both steel are equal and is 3.2 MPa. Carry out a test of hypothesis to determine if the mean yield strength of steels from supplier A is 36.35 MPA at 5% significant level. Determine the 98% confidence interval on the variance ratio between both steels. Also carry out a test of hypothesis on the difference in population mean between both steels to determine if this difference is equal to 2.31 MPa at the 5% sgnificant level. What is the P-Value? c. d. e.

Explanation / Answer

Solution

Let X1 = Yield strength (MPa) of supplier A and X2 = Yield strength (MPa) of supplier B

We X1 ~ N(µ1, ?12) and X2 ~ N(µ2, ?22)

Under the assumption population variances are equal, ?12 = ?12 = ?12.

Preparatory Work

Supplier

(i)

Sample Size

(ni)

Sample Average (Xibar)

Sample Variance (si2)

Sample Standard Deviation (si)

39.6533

21.2443

4.6091

37.3333

14.3086

3.7827

Pooled estimate of ?2 , s2 = 17.7764, s = 4.2162

Part (a)

100(1 - ?) % Confidence Interval for (?1 - ?2) is: (X1bar – X2bar) ± {(t2n – 2, ?/2)(s)?(2/n)}, when population variances are equal but unknown and sample sizes are equal.

So, 98% CI for difference in population means

= (39.6533 – 37.3333) ± {(2.583)(4.2162)?(2/9)}

= 2.32 ± 5.1338

= (- 2.81, 7.45) ANSWER

Part (b)

100(1 - ?) % Confidence Interval for (?1 - ?2) is: (X1bar – X2bar) ± {(Z ?/2)(?)?(2/n)}, when population variances are equal and known and sample sizes are equal.

So, 98% CI for difference in population means

= (39.6533 – 37.3333) ± {(2.325)(3.2)?(2/9)}

= 2.32 ± 3.5072

= (- 1.18, 5.82) ANSWER

Part (c)

Hypotheses:

Null H0: µ = µ0 = 36.35 Vs Alternative HA: µ ? µ0

Test statistic:

t = (?n)(Xbar - µ0)/s, where n = sample size; Xbar = sample average; s = sample standard deviation.

= 3(39.6533 – 36.35)/4.6091

= 2.1501

Distribution, Critical Value and p-value

Under H0, t ~ tn – 1 [i.e., t-distribution with (n -1) degrees of freedom)

Critical value = upper (2.5)% point of t8 = 2.306.

p-value = P(tn – 1 > |2.1501|) = 0.0638

Decision:

Since | tcal | < tcrit, H0 is accepted.

Since p-value > ?. H0 is accepted.

Conclusion:

There is sufficient evidence to support the claim that the mean of supplier A is 36.35.

DONE

ANSWER

Part (d)

100(1 - ?) % Confidence Interval for ?12/ ?22 is:

{(s12/ s22)/ (Fn – 1, n – 1,?/2)}, {(s12/ s22) /(Fn – 1, n – 1, 1 -?/2)}

So, 98% CI is:[{(21.2445/14.3085)/6.0289}, {(21.2445/14.3085)/0.1659}]

= [0.5839, 21.2176] ANSWER