## MATH1005 Quizzes

Quiz 11: Confidence intervals
Question 1 Questions
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) $375±2.756×\frac{81}{\sqrt{30}}$. b) $375±2.575×\frac{9}{\sqrt{30}}$ c) $375±2.33×\frac{81}{\sqrt{30}}$ d) $375±2.575×\frac{81}{\sqrt{30}}$

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 $\stackrel{̄}{x}±z\frac{\sigma }{\sqrt{n}}$, where $\sigma =81$, $n=30$ and where $z$ is chosen to ensure that $P\left(|Z|\le z\right)=0.99.$ From the normal tables, $P\left(|Z|\le 2.575\right)=0.99$ (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) $375±2.756×\frac{91.5}{\sqrt{30}}$ b) $375±2.750×\frac{91.5}{\sqrt{30}}$ c) $375±2.462×\frac{91.5}{\sqrt{30}}$ d) $375±2.575×\frac{91.5}{\sqrt{30}}$

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 $\stackrel{̄}{x}±t\frac{s}{\sqrt{n}}$, where $s=91.5$, $n=30$ and $t=2.756$, since $P\left(|{t}_{29}|\le 2.756\right)=0.99.$.
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)
 a) $19.92±2.81$. b) $19.92±3.05$ c) $19.92±3.04$ d) $19.92±2.51$.

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 $\stackrel{̄}{x}±t\frac{s}{\sqrt{n}}$, where $s=5.73$, $n=16$ and $t=2.131$. (Note that $P\left(|{t}_{15}|\le 2.131\right)=0.95$.)
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)
 a) (-0.21, 0.21) b) (-0.66, 0.25) c) (-0.76, 0.35) d) (-0.63, 0.43)

Choice (a) is incorrect
Try again
Choice (b) is correct!
The sample mean difference is $\stackrel{̄}{d}=-\frac{3.5}{17}=-0.2059$ 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 $-0.2059±1.746×\frac{1.07266}{\sqrt{17}}$. (Note that $P\left(|{t}_{16}|<1.746\right)=0.9$.) 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)
 a) $n=1245$ b) $n=34$. c) $n=95$ d) $n=139$.

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 $2×1.96\frac{3}{\sqrt{n}}$, 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 $\stackrel{̄}{x}±1.96\frac{3}{\sqrt{n}}$. This is of length $2×1.96\frac{3}{\sqrt{n}}$. We need to find the smallest integer $n$ such that $2×1.96\frac{3}{\sqrt{n}}\le 1$, ie $\sqrt{n}\ge 3.92×3$. 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) $0.6±1.96\sqrt{\frac{0.6×0.4}{100}}$ b) $0.6±1.645\sqrt{\frac{0.6×0.4}{100}}$ c) $0.6±1.96$ d) $0.6±1.96\sqrt{\frac{1}{400}}$

Choice (a) is correct!
The approximate 95% confidence interval is $\stackrel{̂}{p}±1.96\sqrt{\frac{p\left(1-p\right)}{n}}$ where $p$ is estimated by $\stackrel{̂}{p}=0.6$. (Note that $P\left(|Z|\le 1.96\right)=0.95$.)
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) $0.6±1.96\sqrt{\frac{1}{400}}$ b) $0.6±1.645\sqrt{\frac{1}{400}}$ c) $0.6±1.645\sqrt{\frac{0.6×0.4}{n}}$ d) $0.6±1.645$

Choice (a) is incorrect
Note that $P\left(|Z|\le 1.96\right)=0.95$, so your interval is a conservative 95% confidence interval.
Choice (b) is correct!
The conservative 90% confidence interval is $\stackrel{̂}{p}±1.645\sqrt{\frac{0.5×0.5}{n}}$ where $\stackrel{̂}{p}=0.6$ and $n=100$ . Note that $P\left(|Z|\le 1.645\right)=0.90$.
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)
 a) There is not enough information to answer this question. b) No, because $n$, the sample size, is too small. c) No, because the length of the confidence interval would be greater than 0.1. d) Yes.

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 $\stackrel{̂}{p}±1.96\sqrt{\frac{1}{400}}$ ie $\stackrel{̂}{p}±0.098$ 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)
 a) $n\approx 96$ b) $n\approx 384$ c) There is not enough information to answer this question. d) $n\approx 1000$.

Choice (a) is incorrect
This would give an interval of length 0.2.
Choice (b) is correct!
You want an interval of the form $\stackrel{̂}{p}±0.05$, 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)
 a) $141 $±$$2.13. b) $141 $±$$2.12. c) $141 $±$$1.96. d) $141 $±$$1.56.

Choice (a) is correct!
Use the ${t}_{15}$ distribution since $\sigma$ is unknown. The 95% confidence interval for the mean is $141±2.131\frac{4}{\sqrt{16}}$ 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 $\sqrt{4}$.