The diameter of steel rods manufactured on two different extrusion machines is being investigated. Two random samples of sizes n1"=15 and n2"=17 are selected, and the sample means and sample variances are x1 =8.73, s2=0.35, x =8.68, and s2=0.40, respectively. Assume that σ1^2 = σ2^2 that the data are drawn from a normal distribution. Required:
a. Is there evidence to support the claim that the two machines produce rods with different mean diameters? Use alpha=0.05 in arriving at this conclusion.
b. Find the P-value for thet-statistic you calculated in part (a).
c. Construct a 95% confidence interval for the difference in mean rod diameter. Interpret this interval.

a) No sufficient evidence to support the claim that the two machines produce rods with different mean diameters.

b) P-value is 0.80

c)  −0.3939 <μ< 0.4939

Step-by-step explanation:

Given Data:

sample sizes

n1 = 15

n2 = 17

sample means:

x1 = 8.73

x2 = 8.68

sample variances:

s1² = 0.35

s2² = 0.40

Hypothesis:

H₀ : μ₁ = μ₂

H₁ :  μ₁ ≠ μ₂

Compute the pooled standard deviation:

= 0.613732

= 0.6137

Compute the test statistic:

= 0.05 / 0.217401

= 0.22999

t = 0.230

Compute degree of freedom:

df = n1 + n2 -2 = 15 + 17 - 2 = 30

Compute the P-value from table using df = 30

P > 2 * 0.40 = 0.80

P > 0.05 ⇒ Fail to reject H₀

Null hypothesis is rejected when P-value is less than or equals to level of significance. But here the P-value = 0.80 and level of significance = 0.05. So P-value is greater than significance level. Hence there is not sufficient evidence to support the claim that population means are different.

Construct a 95% confidence interval for the difference in mean rod diameter:

confidence = c = 95% = 0.95

α = 1 - c

  = 1 - 0.95

α = 0.05

Compute degree of freedom:

df = n1 + n2 -2 = 15 + 17 - 2 = 30

Compute with df = 30 using table:

t₀.₀₂₅ = 2.042

Compute confidence interval:

=

= (8.73 - 8.68) -  2.042 ( 0.6137 )

= 0.05 - 2.042 ( 0.6137 )

= 0.05 - 1.253175

= 0.05 - 1.253175

= 0.05 - 1.253175 (0.35424))

= 0.05 - 0.443925

= −0.393925

= −0.3939

= (8.73 - 8.68) +  2.042 ( 0.6137 )

= 0.05 + 2.042 ( 0.6137 )

= 0.05 + 1.253175

= 0.05 + 1.253175

= 0.05 + 1.253175 (0.35424))

= 0.05 + 0.443925

= 0.493925

= 0.4939

−0.3939 <μ₁ - μ₂< 0.4939

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