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Worked solution

Compare tests before and after pooling two samples

A manufacturing process produces ball bearings with diameters with known standard deviation 0.04 cm. Under normal circumstances, the manufacturing process will produce ball bearings of mean diameter 0.5 cm.

(i) During a routine quality control check, a random sample of 25 ball bearings gives a mean of 0.51 cm. Is there evidence to believe that the manufacturing process is producing ball bearings of the stated diameter? Perform an appropriate test at 5% level of significance. State a necessary assumption for the test to be valid. marks

An enhancement on the manufacturing process will ensure that the diameters of the ball bearings produced are less variable.

(ii) Measurements of a sample of 100 ball bearings give the following summary statistics:

∑x=50.6,∑(x−0.5)2=0.08345.\sum x=50.6,\qquad\sum(x-0.5)^2=0.08345.

Show that the unbiased estimate of the population variance is 8.07×10−48.07\times 10^{-4}.

Is there evidence at the 5% level of significance that after the enhancement, the manufacturing process is producing oversized ball bearings? marks

(iii) Another sample of 100 ball bearings yield the same summary statistics as the previous sample in (ii). Explain, with justification, whether the combined sample will give a different conclusion to (ii). marks

[2017/NYJC/II/9]

CJC · Review problems · Question 7

Answer

(i) z=1.25z=1.25, p=0.211p=0.211: do not reject; assume normal diameters. (ii) s2≈8.07×10−4s^2\approx 8.07\times 10^{-4}, p≈0.0173p\approx 0.0173: reject. (iii) pooled p≈0.00137p\approx 0.00137: reject again.

(i) Test with known population standard deviation

Let XX be an original-process bearing diameter in cm, with mean μ\mu and known σ=0.04\sigma=0.04. Test H0:μ=0.5H_0:\mu=0.5 against H1:μ≠0.5H_1:\mu\ne 0.5. Assume the population diameter distribution is normal, as required for this n 25 exact normal-mean test. Under H0H_0, Xˉ∼N(0.5,0.042/25)\bar X\sim N(0.5,0.04^2/25). Then z=(0.51−0.5)/(0.04/25)=1.25z=(0.51-0.5)/(0.04/\sqrt{25})=1.25 and two-sided p=0.2112995473>0.05p=0.2112995473>0.05. Equivalently, ∣z∣<1.959963985|z|<1.959963985. Do not reject: there is insufficient evidence of a departure from 0.5 cm, not proof of equality. Source Z-Test Stats input: null.5, known SD.04, sample mean.51,n 25, not-equal alternative; outputz 1.25,p.2112996779.

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(ii) Estimate the enhanced-process variance

Let YY be an enhanced-process bearing diameter. Here yˉ=50.6/100=0.506\bar y=50.6/100=0.506 and ∑(y−0.5)=0.6\sum(y-0.5)=0.6. Therefore s2=0.08345−0.62/10099=0.0008065656566≈8.07×10−4.\begin{aligned}&s^2\\ & =\frac{0.08345-0.6^2/100}{99}\\ & =0.0008065656566\approx 8.07\times 10^{-4}.\end{aligned}

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(ii) Test for oversized bearings, both methods

Test H0:μ=0.5H_0:\mu=0.5 against H1:μ>0.5H_1:\mu>0.5, now for the enhanced population. Under H0H_0, Yˉ≈N(0.5,s2/100)\bar Y\approx N(0.5,s^2/100) by the large-sample approximation. Using unrounded s2s^2, z=2.112668648z=2.112668648 and right-tail p=0.01731456927<0.05p=0.01731456927<0.05. Alternatively z>1.644853627z>1.644853627. Reject: evidence at 5% of oversized bearings. The source uses rounded s2=0.000807s^2=0.000807, inputs estimated SD≈.02840745426, mean.506,n 100, null.5, right-tail, and displaysz 2.11210031,p.0173388734; this yields the same conclusion.

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(iii) Pool the two samples

Assuming the additional sample is independent and from the same enhanced process, the pooled sample has n 200, ∑y=101.2\sum y=101.2, yˉ=0.506\bar y=0.506, ∑(y−0.5)=1.2\sum(y-0.5)=1.2 and squared-deviation sum 0.1669. Thus s2=0.1669−1.22/200199=0.0008025125628.\begin{aligned}&s^2\\ & =\frac{0.1669-1.2^2/200}{199}\\ & =0.0008025125628.\end{aligned} Under H0H_0, Yˉ≈N(0.5,s2/200)\bar Y\approx N(0.5,s^2/200); the right-tail statistic is z=2.995300013z=2.995300013 and p=0.001370875126<0.05p=0.001370875126<0.05. Equivalently z>1.644853627z>1.644853627. Reject again: the conclusion stays the same, with stronger evidence. Source calculator inputSD≈.02832866040, mean.506,n 200, null.5, right-tail displaysz≈2.995299197,p≈.001370948. Its prose instead prints.0137; that decimal-place discrepancy is recorded below.

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Original process: 5% two-sided: standard normal, two-tail rejection region, critical magnitude 1.95996, total shaded probability 0.05 (0.025 each tail). Observed statistic 1.25000. Enhanced process: 5% upper-tail: standard normal, right-tail rejection region, critical magnitude 1.64485, total shaded probability 0.05. Observed statistic 2.11267. Pooled samples: 5% upper-tail: standard normal, right-tail rejection region, critical magnitude 1.64485, total shaded probability 0.05. Observed statistic 2.99530.
Original process: 5% two-sided: standard normal, two-tail rejection region, critical magnitude 1.95996, total shaded probability 0.05 (0.025 each tail). Observed statistic 1.25000. Enhanced process: 5% upper-tail: standard normal, right-tail rejection region, critical magnitude 1.64485, total shaded probability 0.05. Observed statistic 2.11267. Pooled samples: 5% upper-tail: standard normal, right-tail rejection region, critical magnitude 1.64485, total shaded probability 0.05. Observed statistic 2.99530.
Original process: 5% two-sided: standard normal, two-tail rejection region, critical magnitude 1.95996, total shaded probability 0.05 (0.025 each tail). Observed statistic 1.25000. Enhanced process: 5% upper-tail: standard normal, right-tail rejection region, critical magnitude 1.64485, total shaded probability 0.05. Observed statistic 2.11267. Pooled samples: 5% upper-tail: standard normal, right-tail rejection region, critical magnitude 1.64485, total shaded probability 0.05. Observed statistic 2.99530.

Common mistakes

For n 25 with known SD, normality is the source’s necessary assumption; “n is large” is not the justification here.

Solution notes

Source part (i) says “n is large” and CLT despite n 25, while also stating normality; the student derivation uses that normality assumption. Source critical 1.6445 should be 1.644853627. In(iii), calculatorp≈.001370948 but prosep=.0137; correct decimal placement and both source versions retained. Variable X/Y slips are reconciled to the enhanced-process variable.

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