Questions 1 and 2 use the same information.

A random sample of 30 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) was recorded for each household in the sample for the March quarter of 2006. The average usage was found to be 375kWh. In a very large study in the March quarter of the previous year it was found that the standard deviation of the usage was 81kWh.

Assuming the standard deviation is unchanged and that the usage is normally distributed, provide an expression for calculating a 99% confidence interval for the mean usage in the March quarter of 2006. Exactly one option must be correct)

A random sample of 30 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) was recorded for each household in the sample for the March quarter of 2006. The average usage was found to be 375kWh. In a very large study in the March quarter of the previous year it was found that the standard deviation of the usage was 81kWh.

Assuming the standard deviation is unchanged and that the usage is normally distributed, provide an expression for calculating a 99% confidence interval for the mean usage in the March quarter of 2006. Exactly one option must be correct)

*Choice (a) is incorrect*

$\sigma $
is assumed known, so the normal tables should be used.

*Choice (b) is incorrect*

The standard deviation
is assumed to be $\sigma =81$.

*Choice (c) is incorrect*

You have given a 98%
confidence interval.

*Choice (d) is correct!*

Since $\sigma $ is assumed known,
we use the interval $\stackrel{\u0304}{x}\pm z\frac{\sigma}{\sqrt{n}}$,
where $\sigma =81$,
$n=30$ and where
$z$ is chosen to ensure
that $P\left(\left|Z\right|\le z\right)=0.99.$ From the
normal tables, $P\left(\left|Z\right|\le 2.575\right)=0.99$
(because $P\left(Z<2.575\right)=0.995$) and
so we use $z=2.575$.

Questions 1 and 2 use the same information.

A random sample of 30 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) was recorded for each household in the sample for the March quarter of 2006. The average usage was found to be 375kWh. In a very large study in the March quarter of the previous year it was found that the standard deviation of the usage was 81kWh.

It is believed that the standard deviation may have changed from the previous year. From the small data set in 2006, the sample standard deviation is 91.5kWh. Assuming that the usage is normally distributed, provide an expression for calculating a 99% confidence interval for the mean usage in the March quarter of 2006. Exactly one option must be correct)

A random sample of 30 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) was recorded for each household in the sample for the March quarter of 2006. The average usage was found to be 375kWh. In a very large study in the March quarter of the previous year it was found that the standard deviation of the usage was 81kWh.

It is believed that the standard deviation may have changed from the previous year. From the small data set in 2006, the sample standard deviation is 91.5kWh. Assuming that the usage is normally distributed, provide an expression for calculating a 99% confidence interval for the mean usage in the March quarter of 2006. Exactly one option must be correct)

*Choice (a) is correct!*

Since
$\sigma $
may have changed from the previous year, we assume
$\sigma $ is unknown, and we
use the ${t}_{29}$ tables. The
confidence interval is $\stackrel{\u0304}{x}\pm t\frac{s}{\sqrt{n}}$,
where $s=91.5$,
$n=30$ and
$t=2.756$, since
$P\left(\left|{t}_{29}\right|\le 2.756\right)=0.99.$.

*Choice (b) is incorrect*

You have used
the wrong $t$
distribution.

*Choice (c) is incorrect*

You have given a 98% confidence interval.

*Choice (d) is incorrect*

Since
$\sigma $ is
unknown, the normal tables should not be used.

An industrial designer wants to determine the average amount of time it takes an
adult to assemble an “easy to assemble” toy. A sample of 16 times yielded an
average time of 19.92 minutes, with a sample standard deviation of 5.73
minutes. Assuming normality of assembly times, provide a 95% confidence
interval for the mean assembly time. Exactly one option must be correct)

*Choice (a) is incorrect*

Since
$\sigma $
is unknown, the normal tables should not be used.

*Choice (b) is correct!*

Since
$\sigma $ is unknown, we use
the ${t}_{15}$ tables, and the
confidence interval is $\stackrel{\u0304}{x}\pm t\frac{s}{\sqrt{n}}$,
where $s=5.73$,
$n=16$ and
$t=2.131$. (Note
that $P\left(\left|{t}_{15}\right|\le 2.131\right)=0.95$.)

*Choice (c) is incorrect*

You have either made a numerical error or you have used the wrong
$t$-distribution.

*Choice (d) is incorrect*

This
is a 90% confidence interval.

A topic of interest in ophthalmology is whether or not spherical refraction differs between
the left and right eye on average. In a study to investigate this, refraction was measured
on the left and right eye of 17 patients. The differences (right - left) in diopters were
${d}_{1},{d}_{2},\cdots {d}_{17}$ and elementary
calculations gave ${\sum}_{i=1}^{17}{d}_{i}=-3.50$
and ${\sum}_{i=1}^{17}{d}_{i}^{2}=19.13$. Provide
a 90% confidence interval (to 2dp) for the average difference (right - left). Exactly one
option must be correct)

*Choice (a) is incorrect*

Try again

*Choice (b) is correct!*

The sample mean difference is
$\stackrel{\u0304}{d}=-\frac{3.5}{17}=-0.2059$ and the sample
variance is ${s}_{d}^{2}=\frac{1}{16}\left(19.13-\frac{{\left(-3.5\right)}^{2}}{17}\right)=1.15059$,
so that ${s}_{d}=1.07266.$
The 90% confidence interval for the difference uses the
${t}_{16}$ distribution,
and is therefore $-0.2059\pm 1.746\times \frac{1.07266}{\sqrt{17}}$.
(Note that $P\left(\left|{t}_{16}\right|<1.746\right)=0.9$.)
Rounding to 2dp gives the answer.

*Choice (c) is incorrect*

This is a 95% confidence
interval.

*Choice (d) is incorrect*

The normal tables should not be used, since the sample standard
deviation of differences is used to obtain the confidence interval.

What is the smallest sample size required to provide a 95% confidence
interval for a mean, if it important that the interval be no longer than
1cm? You may assume that the population is normal with variance
9cm${}^{2}$. Exactly one option
must be correct)

*Choice (a) is incorrect*

$\sigma =3$.
Perhaps you used the value of the variance by mistake.

*Choice (b) is incorrect*

The length of
the interval is $2\times 1.96\frac{3}{\sqrt{n}}$,
not $1.96\frac{3}{\sqrt{n}}$.

*Choice (c) is incorrect*

Try
again.

*Choice (d) is correct!*

Since $\sigma $
is known to be 3cm, a 95% confidence interval is
$\stackrel{\u0304}{x}\pm 1.96\frac{3}{\sqrt{n}}$. This is of length
$2\times 1.96\frac{3}{\sqrt{n}}$. We need to find
the smallest integer $n$
such that $2\times 1.96\frac{3}{\sqrt{n}}\le 1$, ie
$\sqrt{n}\ge 3.92\times 3$. The smallest
$n$ satisfying
this is $n=139$.

Questions 6 and 7 use the same information.

A random sample of 100 preschool children in Camperdown revealed that only 60 had been vaccinated.

Provide an approximate 95% confidence interval for the proportion vaccinated in that suburb. Exactly one option must be correct)

A random sample of 100 preschool children in Camperdown revealed that only 60 had been vaccinated.

Provide an approximate 95% confidence interval for the proportion vaccinated in that suburb. Exactly one option must be correct)

*Choice (a) is correct!*

The approximate 95%
confidence interval is $\widehat{p}\pm 1.96\sqrt{\frac{p\left(1-p\right)}{n}}$
where $p$ is
estimated by $\widehat{p}=0.6$.
(Note that $P\left(\left|Z\right|\le 1.96\right)=0.95$.)

*Choice (b) is incorrect*

This is an approximate
90% confidence interval.

*Choice (c) is incorrect*

Try again.

*Choice (d) is incorrect*

This is a conservative 95% interval.

Questions 6 and 7 use the same information.

A random sample of 100 preschool children in Camperdown revealed that only 60 had been vaccinated.

Provide a conservative 90% confidence interval for the proportion vaccinated in that suburb. Exactly one option must be correct)

A random sample of 100 preschool children in Camperdown revealed that only 60 had been vaccinated.

Provide a conservative 90% confidence interval for the proportion vaccinated in that suburb. Exactly one option must be correct)

*Choice (a) is incorrect*

Note
that $P\left(\left|Z\right|\le 1.96\right)=0.95$,
so your interval is a conservative 95% confidence interval.

*Choice (b) is correct!*

The conservative 90%
confidence interval is $\widehat{p}\pm 1.645\sqrt{\frac{0.5\times 0.5}{n}}$
where $\widehat{p}=0.6$
and $n=100$ . Note
that $P\left(\left|Z\right|\le 1.645\right)=0.90$.

*Choice (c) is incorrect*

This
is an approximate 90% confidence interval.

*Choice (d) is incorrect*

Try
again.

In exploring possible sites for a convenience store in a large neighbourhood, the retail
chain wants to know the proportion of ratepayers in favour of the proposal. If the
estimate is required to be within 0.1 of the true proportion, would a random sample of
size $n=100$
from the council records be sufficient for a 95% confidence interval of this precision?
Exactly one option must be correct)

*Choice (a) is incorrect*

Use a conservative confidence interval.

*Choice (b) is incorrect*

Try again.

*Choice (c) is incorrect*

If the estimate is required to be within 0.1 of
the true proportion, the length of the interval must not exceed 0.2.

*Choice (d) is correct!*

If
$n=100$,
the conservative (longest) 95% interval is of the form
$\widehat{p}\pm 1.96\sqrt{\frac{1}{400}}$ ie
$\widehat{p}\pm 0.098$ which
satisfies the requirements.

To obtain an estimate of the proportion of ‘full time’ university students who have a
part time job in excess of 20 hours per week, the student union decides to
interview a random sample of full time students. They want the length of
their 95% confidence interval to be no greater than 0.1. What size sample,
$n$
should be taken? Exactly one option must be correct)

*Choice (a) is incorrect*

This would give an interval of length 0.2.

*Choice (b) is correct!*

You want an interval
of the form $\widehat{p}\pm 0.05$, so
you need to find $n$
satisfying $1.96\sqrt{\frac{1}{4n}}\approx 0.05$.
Solving gives $n\approx 384$.

*Choice (c) is incorrect*

Use a conservative
confidence interval.

*Choice (d) is incorrect*

Try
again.

The recommended retail price of a brand of designer jeans is $150. The price of the
jeans in a sample of 16 retailers is on average $141 with a sample standard deviation of
$4$. If this is a
‘random’ sample and the prices can be assumed to be normally distributed, construct a 95%
confidence interval for the average sale price. Exactly one option must be correct)

*Choice (a) is correct!*

Use the
${t}_{15}$ distribution
since $\sigma $
is unknown. The 95% confidence interval for the mean is
$141\pm 2.131\frac{4}{\sqrt{16}}$ as required.

*Choice (b) is incorrect*

You have used
the wrong $t$-distribution.

*Choice (c) is incorrect*

$\sigma $
is unknown, so the normal distribution should not be used.

*Choice (d) is incorrect*

The sample standard
deviation is $4, not $\sqrt{\$4}$.