Question

Mist (airborne droplets or aerosols) is generated when metal-removing fluids are used in machining operations to cool and lubricate the tool and workpiece. Mist generation is a concern to OSHA, which has recently lowered substantially the workplace standard. An article gave the accompanying data on x = fluid-flow velocity for a 5% soluble oil (cm/sec) and y = the extent of mist droplets having diameters smaller than 10 µm (mg/m3): x 90 177 191 354 360 442 964 y 0.38 0.60 0.45 0.66 0.60 0.69 0.97 (a) The investigators performed a simple linear regression analysis to relate the two variables. Does a scatter plot of the data support this strategy? Yes, a scatter plot shows a reasonable linear relationship. No, a scatter plot does not show a reasonable linear relationship. Correct: Your answer is correct. (b) What proportion of observed variation in mist can be attributed to the simple linear regression relationship between velocity and mist? (Round your answer to three decimal places.) Incorrect: Your answer is incorrect. (c) The investigators were particularly interested in the impact on mist of increasing velocity from 100 to 1000 (a factor of 10 corresponding to the difference between the smallest and largest x values in the sample). When x increases in this way, is there substantial evidence that the true average increase in y is less than 0.6? (Use α = 0.05.) State the appropriate null and alternative hypotheses. H0: β1 = 0.0006667 Ha: β1 ≠ 0.0006667 H0: β1 = 0.0006667 Ha: β1 > 0.0006667 H0: β1 = 0.0006667 Ha: β1 < 0.0006667 H0: β1 ≠ 0.0006667 Ha: β1 = 0.0006667 Correct: Your answer is correct. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = Incorrect: Your answer is incorrect. P-value = Incorrect: Your answer is incorrect. State the conclusion in the problem context. Fail to reject H0. There is not sufficient evidence that with an increase from 100 to 1000, the true average increase in y is less than 0.6. Fail to reject H0. There is sufficient evidence that with an increase from 100 to 1000, the true average increase in y is less than 0.6. Reject H0. There is sufficient evidence that with an increase from 100 to 1000, the true average increase in y is less than 0.6. Reject H0. There is not sufficient evidence that with an increase from 100 to 1000, the true average increase in y is less than 0.6. Correct: Your answer is correct. (d) Estimate the true average change in mist associated with a 1 cm/sec increase in velocity, and do so in a way that conveys information about precision and reliability. (Calculate a 95% CI. Round your answers to six decimal places.) Incorrect: Your answer is incorrect. , Incorrect: Your answer is incorrect. mg/m3

Answer 1

(a)

The scatter plot for the given data is,

The scatter plot shows a linear relation between x and y. Thus,

Yes, a scatter plot shows a reasonable linear relationship.

b)

= 314.0071² / (506045.4 * 0.2162857)

= 0.901

So, the proportion of observed variation in mist can be attributed to the simple linear regression relationship between velocity and mist is 0.901

c)

The the appropriate null and alternative hypotheses.

H0: β1 = 0.0006667

Ha: β1 < 0.0006667

Estimated value of is b1 = Sxy / Sxx = 314.0071 / 506045.4 = 0.0006205117

Sum of squares error, SSE = Syy - Sxy² / Sxx

= 0.2162857 - 314.0071² / 506045.4

= 0.02144062

Standard error of the estimate, se =

Standard error of = se /

= 0.06548377 /

= 0.00009205321

Test statistic, t = b1 / Standard error of

= (0.0006205117 - 0.0006667) / 0.00009205321

= -0.50

Degree of freedom = n-2 = 7 - 2 = 5

P-value = P(t < -0.50) = 0.319

As, p-value is greater than significance level of 0.05,

Fail to reject H0. There is not sufficient evidence that with an increase from 100 to 1000, the true average increase in y is less than 0.6.

d)

With 1 cm/sec increase in velocity, the true average change in mist is 0.000621 mg/m3 (The value of the slope b1)

For df = 5, t value at 95% CI is 2.57

95% CI is,

(0.000621 - 2.57 * 0.00009205321, 0.000621 + 2.57 * 0.00009205321)

(0.000384, 0.000858)