Confidence interval for a mean | CAIE 9709 Probability & Statistics 2 Notes

Probability & Statistics 2

Mathematics9709

Estimate plus or minus a critical margin

\bar x\pm k\frac{\sigma}{\sqrt n}

Symbols

$\bar x$ = sample mean (same as x)
$k$ = central normal critical value (none)
$\sigma/\sqrt n$ = standard error (same as x)

The observations must come from a random sample.
Use $s/\sqrt n$ for a large sample when σ is unknown.
The common k-values are 1.645, 1.960, 2.326 and 2.576 for 90%, 95%, 98% and 99% confidence.
For another level C%, use inverse normal with cumulative area $(1+C/100)/2$ .

Worked example

Construct a 95% interval

A normal population has σ = 6. A sample of 36 has $\bar x=72.4$ .

72.4\pm1.96\left(\frac{6}{\sqrt{36}}\right)

=72.4\pm1.96

Answer

70.44<\mu<74.36

Confidence describes the method, not individual values

Definition

1 mark

What is meant by 95% confidence interval for a population mean?

Model answer: An interval produced by a method that, over repeated random samples, contains the population mean in about 95% of cases.

Different samples, different intervals

fixed μ

sample 1

sample 2

sample 3

sample 4

sample 5

sample 6

sample 7

sample 8

sample 9

sample 10

blue contains μ

red misses μ

Confidence belongs to the long-run method: different samples produce different intervals.

The interval estimates $\mu$ ; it does not contain 95% of individual observations.
A higher confidence level uses a larger k and gives a wider interval.
The width is $2k\sigma/\sqrt n$ . Quadrupling n halves the width.

Reverse a width and compare nested intervals

Worked example

Current-paper confidence-level comparison

An α% interval has 1.414 times the width of a 90% interval from the same sample.

k_{\alpha}=1.414(1.645)=2.326

Answer

\alpha=98.0

Let A mean “the 90% interval contains μ” and B mean “the 98% interval contains μ”. Since A is inside B,

P(A\mid B)=\frac{0.90}{0.98}

Answer

\frac{45}{49}=0.918

(9709/61/O/N/25 Q4)

Your turnIndependent confidence-interval check [4 marks]

A large random sample of 64 has mean 15.2 and sample standard deviation 4.8. Find an approximate 95% confidence interval for the population mean.

Show worked answer

15.2\pm1.96\left(\frac{4.8}{\sqrt{64}}\right)

=15.2\pm1.176

Answer

14.0<\mu<16.4\ \text{(3 s.f.)}

Common mistake

Do not use the full width as the margin. The margin is half the width. Also do not say that there is a 95% chance the fixed parameter lies in a completed interval.

Estimate plus or minus a critical margin

\bar x\pm k\frac{\sigma}{\sqrt n}

Symbols

$\bar x$ = sample mean (same as x)
$k$ = central normal critical value (none)
$\sigma/\sqrt n$ = standard error (same as x)

The observations must come from a random sample.
Use $s/\sqrt n$ for a large sample when σ is unknown.
The common k-values are 1.645, 1.960, 2.326 and 2.576 for 90%, 95%, 98% and 99% confidence.
For another level C%, use inverse normal with cumulative area $(1+C/100)/2$ .

Worked example

Construct a 95% interval

A normal population has σ = 6. A sample of 36 has $\bar x=72.4$ .

72.4\pm1.96\left(\frac{6}{\sqrt{36}}\right)

=72.4\pm1.96

Answer

70.44<\mu<74.36

Confidence describes the method, not individual values

Definition

1 mark

What is meant by 95% confidence interval for a population mean?

Model answer: An interval produced by a method that, over repeated random samples, contains the population mean in about 95% of cases.

Different samples, different intervals

fixed μ

sample 1

sample 2

sample 3

sample 4

sample 5

sample 6

sample 7

sample 8

sample 9

sample 10

blue contains μ

red misses μ

Confidence belongs to the long-run method: different samples produce different intervals.

The interval estimates $\mu$ ; it does not contain 95% of individual observations.
A higher confidence level uses a larger k and gives a wider interval.
The width is $2k\sigma/\sqrt n$ . Quadrupling n halves the width.

Reverse a width and compare nested intervals

Worked example

Current-paper confidence-level comparison

An α% interval has 1.414 times the width of a 90% interval from the same sample.

k_{\alpha}=1.414(1.645)=2.326

Answer

\alpha=98.0

Let A mean “the 90% interval contains μ” and B mean “the 98% interval contains μ”. Since A is inside B,

P(A\mid B)=\frac{0.90}{0.98}

Answer

\frac{45}{49}=0.918

(9709/61/O/N/25 Q4)

Your turnIndependent confidence-interval check [4 marks]

A large random sample of 64 has mean 15.2 and sample standard deviation 4.8. Find an approximate 95% confidence interval for the population mean.

Show worked answer

15.2\pm1.96\left(\frac{4.8}{\sqrt{64}}\right)

=15.2\pm1.176

Answer

14.0<\mu<16.4\ \text{(3 s.f.)}

Common mistake

Do not use the full width as the margin. The margin is half the width. Also do not say that there is a 95% chance the fixed parameter lies in a completed interval.