School of Mathematics and Statistics
Junior
The University of Sydney
Maths & Stats Home / Teaching program / Junior / MATH1015 / Quizzes / Quiz 11
The University of Sydney
Usyd Home
MyUni
MyUni
Library
Sitemap
spcr
   Questions




right first
    attempt
right
wrong

Quiz 11: Confidence intervals

Last unanswered question  Question  Next unanswered question
 

Question 1

 
 
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.
a) 375 ± 2.756 ×√8130.   b) 375 ± 2.575 ×√9- 30
c) 375 ± 2.33 ×√81- 30   d) 375 ± 2.575 ×-81 √30

 

Not correct. Choice (a) is false.
σ is assumed known, so the normal tables should be used.
Not correct. Choice (b) is false.
The standard deviation is assumed to be σ = 81.
Not correct. Choice (c) is false.
You have given a 98% confidence interval.
Your answer is correct.
Since σ is assumed known, we use the interval ¯x ± z-σ- √n, where σ = 81, n = 30 and where z is chosen to ensure that P(|Z|≤ z) = 0.99. From the normal tables, P(|Z|≤ 2.575) = 0.99 (because P(Z < 2.575) = 0.995) and so we use z = 2.575.
 

Question 2

 
 
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.
a) 375 ± 2.756 ×9√1.5 30   b) 375 ± 2.750 ×91.5 √30
c) 375 ± 2.462 ×91.5 √30   d) 375 ± 2.575 ×91.5 √30-

 

Your answer is correct.
Since σ may have changed from the previous year, we assume σ is unknown, and we use the t29 tables. The confidence interval is ¯x ± ts√-- n, where s = 91.5, n = 30 and t = 2.756, since P(|t29|≤ 2.756) = 0.99..
Not correct. Choice (b) is false.
You have used the wrong t distribution.
Not correct. Choice (c) is false.
You have given a 98% confidence interval.
Not correct. Choice (d) is false.
Since σ is unknown, the normal tables should not be used.
 

Question 3

 
 
An idustrial 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.
a) 19.92 ± 2.81.   b) 19.92 ± 3.05
c) 19.92 ± 3.04   d) 19.92 ± 2.51.

 

Not correct. Choice (a) is false.
Since σ is unknown, the normal tables should not be used.
Your answer is correct.
Since σ is unknown, we use the t15 tables, and the confidence interval is ¯x ± ts√-- n, where s = 5.73, n = 16 and t = 2.131. (Note that P(|t15|≤ 2.131) = 0.95.)
Not correct. Choice (c) is false.
You have either made a numerical error or you have used the wrong t-distribution.
Not correct. Choice (d) is false.
This is a 90% confidence interval.
 

Question 4

 
 
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 d1,d2,⋅⋅⋅d17 and elementary calculations gave ∑17 di = - 3.50 i=1 and 1∑7 d2i = 19.13 i=1 . Provide a 90% confidence interval (to 2dp) for the average difference (right - left).
a) (-0.21, 0.21)   b) (-0.66, 0.25)
c) (-0.76, 0.35)   d) (-0.63, 0.43)

 

Not correct. Choice (a) is false.
Try again
Your answer is correct.
The sample mean difference is ¯d = - 31.57 = - 0.2059 and the sample variance is s2d = 116(19.13 - (-3.175)2) = 1.15059 , so that sd = 1.07266. The 90% confidence interval for the difference uses the t16 distribution, and is therefore -0.2059 ± 1.746 ×1.√07266 17. (Note that P(|t16| < 1.746) = 0.9.) Rounding to 2dp gives the answer.
Not correct. Choice (c) is false.
This is a 95% confidence interval.
Not correct. Choice (d) is false.
The normal tables should not be used, since the sample standard deviation of differences is used to obtain the confidence interval.
 

Question 5

 
 
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 9cm2.
a) n = 1245   b) n = 34.
c) n = 95   d) n = 139.

 

Not correct. Choice (a) is false.
σ = 3. Perhaps you used the value of the variance by mistake.
Not correct. Choice (b) is false.
The length of the interval is 2 × 1.96√3- n, not 1.96√3- n.
Not correct. Choice (c) is false.
Try again.
Your answer is correct.
Since σ is known to be 3cm, a 95% confidence interval is ¯x ± 1.963√-- n. This is of length 2 × 1.96-3- √n. We need to find the smallest integer n such that 2 × 1.96√3n- ≤ 1, ie √ -- n ≥ 3.92 × 3. The smallest n satisfying this is n = 139.
 

Question 6

 
 
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.
a) ∘ -0.6×-0.4 0.6± 1.96 -------- 100   b) ∘ 0.6-×-0.4- 0.6± 1.645 -------- 100
c) 0.6 ± 1.96   d) ∘ ---- 0.6 ±1.96 -1- 400

 

Your answer is correct.
The approximate 95% confidence interval is ∘ p(1--p)- ˆp± 1.96 ---n--- where p is estimated by ˆp = 0.6. (Note that P(|Z|≤ 1.96) = 0.95.)
Not correct. Choice (b) is false.
This is an approximate 90% confidence interval.
Not correct. Choice (c) is false.
Try again.
Not correct. Choice (d) is false.
This is a conservative 95% interval.
 

Question 7

 
 
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.
a) ∘ ---- -1- 0.6 ± 1.96 400   b) ∘ ---- -1- 0.6 ±1.645 400
c) ∘ 0.6×-0.4- 0.6 ± 1.645 -------- n   d) 0.6 ± 1.645

 

Not correct. Choice (a) is false.
Note that P(|Z|≤ 1.96) = 0.95, so your interval is a conservative 95% confidence interval.
Your answer is correct.
The conservative 90% confidence interval is -------- ∘ 0.5 × 0.5 ˆp ±1.645 ---n---- where pˆ = 0.6 and n = 100 . Note that P(|Z|≤ 1.645) = 0.90.
Not correct. Choice (c) is false.
This is an approximate 90% confidence interval.
Not correct. Choice (d) is false.
Try again.
 

Question 8

 
 
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?
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.

 

Not correct. Choice (a) is false.
Use a conservative confidence interval.
Not correct. Choice (b) is false.
Try again.
Not correct. Choice (c) is false.
If the estimate is required to be within 0.1 of the true proportion, the length of the interval must not exceed 0.2.
Your answer is correct.
If n = 100, the conservative (longest) 95% interval is of the form ∘--- ˆp± 1.96 1-- 400 ie ˆp± 0.098 which satisfies the requirements.
 

Question 9

 
 
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?
a) n ≈ 96
b) n ≈ 384
c) There is not enough information to answer this question.
d) n ≈ 1000.

 

Not correct. Choice (a) is false.
This would give an interval of length 0.2.
Your answer is correct.
You want an interval of the form ˆp± 0.05, so you need to find n satisfying ∘ --- 1 1.96 4n-≈ 0.05 . Solving gives n ≈ 384.
Not correct. Choice (c) is false.
Use a conservative confidence interval.
Not correct. Choice (d) is false.
Try again.
 

Question 10

 
 
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.
a) $141 ± $2.13.   b) $141 ± $2.12.
c) $141 ± $1.96.   d) $141 ± $1.56.

 

Your answer is correct.
Use the t15 distribution since σ is unknown. The 95% confidence interval for the mean is 141 ± 2.131√416 as required.
Not correct. Choice (b) is false.
You have used the wrong t-distribution.
Not correct. Choice (c) is false.
σ is unknown, so the normal distribution should not be used.
Not correct. Choice (d) is false.
The sample standard deviation is $4, not √-- $4.