SI12, SI13 - Confidence intervals from software

SI12, SI13 - Confidence intervals from software

SI12 reads confidence intervals from software, including output for a parameter or distribution whose calculation has been supplied. SI13 uses the reported interval to assess a hypothesised parameter value, with a conclusion that reflects sampling uncertainty.

Read the parameter and limits before making a claim

Can the same data give a 95%95\% interval that excludes a proposed mean and a 99%99\% interval that includes it? Yes. Greater confidence produces a wider interval, so the strength of the conclusion must name the confidence level.

You already know

A confidence interval estimates a fixed population parameter. Its confidence level describes repeated coverage by the method. Standard error concerns the estimate's sampling variability, while standard deviation concerns individual observations.

Read a software extract in this order: identify the population parameter and units, identify the estimate, then read the confidence level and both limits. A label such as “difference” is incomplete until you know which quantity was subtracted from which.

Worked example 1: an interval reported as a margin

Question. Software gives the following large-sample Normal output for independent processing times, in minutes. State and interpret the confidence interval.

FieldValue
Sample size6464
Sample mean12.412.4
Sample standard deviation2.42.4
Standard error0.30.3
Confidence level0.950.95
Interval12.4±0.58812.4\pm0.588

Plan. Convert the centre and margin into two limits, then identify the population quantity being estimated.

L=12.4−0.588=11.812lower confidence limitU=12.4+0.588=12.988upper confidence limit\begin{aligned} L&=12.4-0.588=11.812 && \text{lower confidence limit}\\ U&=12.4+0.588=12.988 && \text{upper confidence limit} \end{aligned}

Why this step? The interval concerns the population mean processing time, not the spread of the individual times.

Answer. A 95%95\% confidence interval for the population mean processing time is (11.812,12.988)(11.812,12.988) minutes.

Check. The midpoint is 12.412.4; the reported standard error is 2.4/64=0.32.4/\sqrt{64}=0.3.

Change only the confidence level to 90%90\%, leaving the sample and estimated standard error unchanged.

Worked example 2: software reports the endpoints instead

Question. The new software output reports:

FieldValue
Mean estimate12.412.4
Standard error0.30.3
Confidence level0.900.90
Lower limit11.906511.9065
Upper limit12.893512.8935

Express the interval as centre plus or minus margin, and explain the change in width.

Plan. Recover the midpoint and half-width from the two reported endpoints.

centre=11.9065+12.89352=12.4midpointmargin=12.8935−11.90652=0.4935half the widthinterval=12.4±0.4935equivalent form\begin{aligned} \text{centre}&=\frac{11.9065+12.8935}{2}=12.4 && \text{midpoint}\\ \text{margin}&=\frac{12.8935-11.9065}{2}=0.4935 && \text{half the width}\\ \text{interval}&=12.4\pm0.4935 && \text{equivalent form} \end{aligned}

Why this step? A margin is half the total width. Software may report either form without changing what the interval estimates.

Answer. 12.4±0.493512.4\pm0.4935 minutes. It is narrower because the confidence level is lower.

Check. 1.645(0.3)=0.49351.645(0.3)=0.4935, consistent with the reported Normal method.

The calculations above interpret the output; they do not require a new software-specific formula. The booklet supplies the mean-interval construction, but does not supply a rule for identifying what a particular output label means.

Watch out

Wrong: use 2.42.4 as the estimate's standard error because it is labelled “standard deviation”.

Right: read the field labels. Here 2.42.4 describes individual times, while 0.30.3 describes the estimated mean's sampling spread.

Read another confidence level for the same sample.

The reported contrast decides what the signs mean

Software can report intervals for differences from independent groups as well as for paired differences. It can also report an interval for a less familiar parameter, with its meaning and necessary details supplied. Interpreting such an interval does not require inventing a construction formula.

Always read the method description. An interval labelled “two independent groups” is not an instruction to manufacture pairs. Conversely, paired measurements should retain their genuine links.

Worked example 3: output for independent groups

Question. Two independent random groups test materials A and B. Software uses a valid method for these independent samples and reports the following 95%95\% confidence interval, in megapascals, for θ=μA−μB\theta=\mu_A-\mu_B:

θ^=0.8,(L,U)=(−0.3,1.9)\hat\theta=0.8,\qquad (L,U)=(-0.3,1.9)

Interpret the estimate and interval.

Plan. Translate the signed difference into a comparison of the two population means.

θ^=0.8estimated A-minus-B mean strengthL+U2=−0.3+1.92=0.8midpoint agreesU−L2=1.9−(−0.3)2=1.1reported margin\begin{aligned} \hat\theta&=0.8 && \text{estimated A-minus-B mean strength}\\ \frac{L+U}{2}&=\frac{-0.3+1.9}{2}=0.8 && \text{midpoint agrees}\\ \frac{U-L}{2}&=\frac{1.9-(-0.3)}2=1.1 && \text{reported margin} \end{aligned}

Why this step? A negative difference means A's population mean is lower; a positive difference means it is higher.

Answer. The estimated mean strength of A exceeds that of B by 0.8 MPa0.8\,\mathrm{MPa}. The 95%95\% interval for their population mean difference runs from A being 0.3 MPa0.3\,\mathrm{MPa} lower to A being 1.9 MPa1.9\,\mathrm{MPa} higher.

Check. The interval crosses zero, so it includes differences in either direction.

Change only the software's contrast from A minus B to B minus A.

Worked example 4: the opposite contrast

Question. Re-express that output for ϕ=μB−μA\phi=\mu_B-\mu_A.

Plan. Negate the estimate and reverse the negated interval endpoints.

ϕ^=−0.8opposite contrast(Lnew,Unew)=(−1.9,0.3)negate and reorder\begin{aligned} \hat\phi&=-0.8 && \text{opposite contrast}\\ (L_{\mathrm{new}},U_{\mathrm{new}})&=(-1.9,0.3) && \text{negate and reorder} \end{aligned}

Why this step? Merely adding a minus sign before both endpoints would leave them in decreasing order.

Answer. The estimate is −0.8 MPa-0.8\,\mathrm{MPa}, with 95%95\% interval (−1.9,0.3) MPa(-1.9,0.3)\,\mathrm{MPa} for B minus A.

Check. It describes exactly the same comparison as the first output, with the labels reversed.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account