Done in excel ( show formulas used) 4.You are assessing two competing lines for

Question

Done in excel ( show formulas used)

4.You are assessing two competing lines for productivity. The Data (after question) provides data for number produced each hour for 48 hours, on each of the two lines of interest. Determine a 95% confidence interval for the difference between the two. What do you conclude?

497.699503 505.7534527 499.925984 505.7249566 500.926015 506.228272 501.565635 509.0091547 501.237301 503.4450333 499.518384 507.9718904 497.590787 504.0480652 501.074532 502.4193895 497.888059 504.3167194 499.762539 505.3011888 502.542325 508.6171062 499.785504 499.7875682 500.414057 506.4966426 500.013271 505.5474331 500.295743 502.0610276 499.144333 506.5136906 500.071643 505.8254968 500.033664 504.6136578 497.103862 506.6775687 501.661784 506.1998215 501.454926 506.8383554 499.084245 505.0106359 498.989194 504.0339587 495.981823 505.4513492 498.707624 506.924895 499.791061 506.5370695 499.644455 504.2568175 495.834864 506.5920175 498.684094 505.6176067 501.359303 506.4576539 501.041118 506.1080647 502.960296 501.5300468 499.068245 502.0412925 500.89295 503.7003172 499.026917 504.5071002 501.162091 503.4411793 503.914596 506.0078421 499.529168 504.6350467 499.334529 504.8189789 500.965844 502.2321988 499.604918 504.5976096 499.457819 504.8178665 496.579425 503.6049023 498.104234 507.0217615 498.929041 504.3959806 500.116783 507.6475165 499.523282 506.1429098 501.660506 506.8457203

Explanation / Answer

Using Minitab:

Two-sample T for C3 vs C4

N Mean StDev SE Mean
C3 48 499.78 1.67 0.24
C4 48 505.17 1.87 0.27

Difference = (C3) - (C4)
Estimate for difference: -5.390
95% CI for difference: (-6.110, -4.669)
T-Test of difference = 0 (vs ): T-Value = -14.86 P-Value = 0.000 DF = 92

95% confidence interval for the difference between the two is  (-6.110, -4.669)

Hope this will helpful. Thanks and God Bless you :)

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