## The diameter of steel rods manufactured on two different extrusion machines is being investigated. Two random samples of sizes n1=15an

The diameter of steel rods manufactured on two different extrusion machines is being investigated. Two random samples of sizes

n1=15andn1=17

are selected, and the sample means and sample variances are

x¯¯¯1=8.73

,

s21=0.35

,

x¯¯¯2=8.68

, and

s22=0.40

, respectively. Assume that

σ21=σ22

and that the data are drawn from a normal distribution. a. Is there evidence to support the claim that the two machines produce rods with different mean diameters? Use α = 0.05 in arriving at this conclusion. Find the P-value. b. Construct a 95% confidence interval for the difference in mean rod diameter. Interpret this interval.

## Answers ( )

Answer:

(a) There is no evidence to support the claim that the two machines produce rods with different mean diameters.

P-value is 0.818

(b) 95% confidence interval for the difference in mean rod diameter is (-0.17, 0.27).

This interval shows that the difference in mean is between -0.17 and 0.27.

Step-by-step explanation:

(a) Null hypothesis: The two machines produce rods with the same mean diameter.

Alternate hypothesis: The two machines produce rods with different mean diameter.

Machine 1

mean = 8.73

variance = 0.35

n1 = 15

Machine 2

mean = 8.68

variance = 0.4

n2 = 17

pooled variance = [(15-1)0.35 + (17-1)0.4] ÷ (15+17-2) = 11.3 ÷ 30 = 0.38

Test statistic (t) = (8.73 – 8.68) ÷ sqrt[0.38(1/15 + 1/17)] = 0.05 ÷ 0.218 = 0.23

degree of freedom = n1+n2-2 = 15+17-2 = 30

Significance level = 0.05 = 5%

Critical values corresponding to 30 degrees of freedom and 5% significance level are -2.042 and 2.042.

Conclusion:

Fail to reject the null hypothesis because the test statistic 0.23 falls within the region bounded by the critical values -2.042 and 2.042.

There is no evidence to support the claim that the two machines produce rods with different mean diameters.

Cumulative area of test statistic is 0.5910

The test is a two-tailed test.

P-value = 2(1 – 0.5910) = 2×0.409 = 0.818

(b) Difference in mean = 8.73 – 8.68 = 0.05

pooled sd = sqrt(pooled variance) = sqrt(0.38) = 0.62

Critical value (t) = 2.042

E = t×pooled sd/√n1+n2 = 2.042×0.62/√15+17 = 0.22

Lower limit of difference in mean = 0.05 – 0.22 = -0.17

Upper limit of difference in mean = 0.05 + 0.22 = 0.27

95% confidence interval for the difference in mean rod diameter is between a lower limit of -0.17 and an upper limit of 0.27.