ResEcon 11

E = za/2 * o/sqrtn is called the

Margin of error

True or False: A sample of 100 fuses from a very large shipment is found to have 10 that are defective. The 0.95 confidence interval would indicate that, for this shipment, the proportion of defective fuses is between 0 and 0.28.

F

The mean and standard deviation for voltages of power packs labeled as 12 volts for a sample of 17 are as follows: xbar=11.57 s=0.3

lower limit : 11.416 upper limit: 11.724

because s is also an estimator (for the population standard deviation), it introduces its own uncertainty due to sampling variation. the sample standard deviation s is a random variable.

subtract/add margin of error to get confidence interval. we can not use the z-distribution to construct an appropriate confidence interval

if we do not know the standard deviation, we can not estimate it because the whole point of interval estimation of the population mean is to account for sampling variation in estimating the population mean. hence we allow for sampling error in estimating the population

we do so by adding and subtracting a margin of error which depends on a pre-assigned level of confidence

once p is computed, sqrt of p(1-p)/n can be computed

without any additional estimation - no degree of freedom lost - no additional variation entered from estimating the standard error

number of observations in that category

x

not accounting for the uncertainty in the estimation of the population standard deviation results in a confidence interval that is not conservative enough. therefore we need to replace the standard normal distribution in

xbar + za/2 * s/sqrtn that is conservative enough to account for the additional error in estimating the standard deviation. we need to replace z

t-distribution is a family of distributions

- depends on degrees of freedom (df) - degrees of freedom (df) depends on the sample size

A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 263 convicts, the official determines that there are 77 cases of recidivism. Find the lower limit of the 80% confidence interval estimate for the population proportion of cases of recidivism.

0.257

A quality control engineer is interested in estimating the proportion of defective items coming off a production line. In a sample of 393 items, 48 are defective. Calculate a 95.0% confidence interval estimate for the proportion of defectives from this production line. (Use 3 decimal places in calculations and in reporting your answers.)

lower limit: 0.09 upper limit: 0.155

for npi >= 10,

n(1-pi) >= 10

for the t distribution, as the sample size (n) increases ,

t--> z (looks more and more like the standard normal)

z score of 90% 95% 99%

1.645 1.96 2.576

A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 250 convicts, the official determines that there are 65 cases of recidivism. A confidence interval will be obtained for the proportion of cases of recidivism. Part of this calculation includes the estimated standard error of the sample proportion. For this sample, the estimated standard error is ----?

0.028

A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 274 convicts, the official determines that there are 113 cases of recidivism. Find the standard error for the population proportion of cases of recidivism.

0.0297

A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 283 convicts, the official determines that there are 39 cases of recidivism. Find the margin of error for a 95% confidence interval estimate for the population proportion of cases of recidivism.

0.0402

Find the margin of error of a poll, assuming 95% confidence, a population proportion of 0.137, and sample size n = 213.

0.05 (A population proportion (π) is the fraction of a population with a certain characteristic or attribute.Though, here we do not know the population standard deviation (σ) we always use a Z value for Proportion. Thus, with our 95% confidence we will use 1.96 for the Z value. E = 1.96 × √((0.137(1-0.137))/213) E = 1.96 × √(0.118/213) E = 1.96 × 0.024 = 0.046 = 0.05)

A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 224 convicts, the official determines that there are 48 cases of recidivism. Find the margin of error for a 95% confidence interval estimate for the population proportion of cases of recidivism.

0.0537

A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 283 convicts, the official determines that there are 62 cases of recidivism. Find the lower limit of the 99% confidence interval estimate for the population proportion of cases of recidivism.

0.1557 0.219 - 2.58*sqrt[(0.219*0.781)/283]

A sample of 16 eggs yields a sample mean weight of 2.00 ounces and a sample standard deviation = 0.360 ounces, find the margin of error in estimating a confidence interval estimate at the 90% level of confidence. Round your answer to three decimal places. You may assume that egg weights are normally distributed

0.158

A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 161 convicts, the official determines that there are 40 cases of recidivism. Find the lower limit of the 95% confidence interval estimate for the population proportion of cases of recidivism.

0.1817

A quality control engineer is interested in estimating the proportion of defective items coming off a production line. In a sample of 303 items, 47 are defective. Calculate a upper confidence limit for a 95.0% confidence interval for the proportion of defectives from this production line.

0.196 (You need the general form for a confidence interval estimate: p +/- za/2 * sqrt of ((p(1-p)/n)) First, calculate the sample proportion: = 47/303 = 0.155.Next, use this value, the appropriate Zα value (1.96) and the standard error, which is determined using the sample proportion (0.155) and the sample size (303). For this problem the lower and upper limits are:0.155 - 1.96*[(0.155*0.845)/303]0.5 = 0.114;and0.155 + 1.96*[(0.155*0.845)/303]0.5 = 0.196You were asked to provide only the upper limit.

calculating the CI for proportions: steps

1. given the confidence level: (1-a)%, calculate a 2. calculate 1-a/2 3. find the z-value corresponding to 1-a/2 (example: for 0.0975, the z-value is 1.96) 4. calculate the CI

The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books checked out each day. The assistant provides the following interval estimate: from 740 to 920 books per day. If the head librarian knows that the population standard deviation is 150 books checked out per day, and she asked her assistant for a 95% confidence interval, approximately how large a sample did her assistant use to determine the interval estimate?

11

A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statistics students generated the following 90% confidence interval: (0.443, 0.637). Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence?

150

A quality control engineer is interested in the mean length of sheet insulation being cut automatically by machine. The desired length of the insulation is 12 feet. It is known that the standard deviation in the cutting length is 0.15 feet. A sample of 70 cut sheets yields a mean length of 12.14 feet. This sample will be used to obtain a 99% confidence interval for the mean length cut by machine.Suppose the engineer had decided to estimate the mean length to within 0.03 with 99% confidence. Then the sample size would be

167 (The margin of error, E, is a function of sample size. As n increases, the E decreases all else held constant. We can use the general formula to find n given a margin of error to work with and a confidence level. The formula is: n= [(Z*population standard deviation)/E]2 round up to next highest integer value for n where: Z is the appropriate critical Z given the confidence level you have been asked to use.In this particular problem: n= [(2.58*.15)/.03]2 =166.41 so round up to 167 for the sample size to use.)

To become an actuary, it is necessary to pass a series of 10 exams, including the most important one, an exam in probability and statistics. An insurance company wants to estimate the mean score on this exam for actuarial students who have enrolled in a special study program. They have no other information about scores, so they take a sample of 6 actuarial students in this program and determine that their scores are: 5, 8, 7, 6, 5, and 7. This sample will be used to calculate a 90% confidence interval for the mean score for actuarial students in the special study program. The critical value used in constructing a 90% confidence interval is

2.015

Suppose a random variable has population mean 131 and population standard deviation 15.9. What is the standard error of the sample mean of a sample of 41 observations?

2.48 (o/sqrtn 15.9/√41 = 2.483 = 2.48)

If you were constructing a 99% confidence interval of the population mean based on a sample of n=25 where the standard deviation of the sample s = 0.05, the critical value of t will be

2.7969

The actual voltages of power packs labeled as 12 volts are as follows: 11.77, 11.90, 11.64, 11.84, 12.13, 11.99, 11.77. The critical value for a 99% confidence interval for this sample is

3.707 (Go to the t-table. Along the top row, locate t with the subscript 0.005. Along the left column, under df, locate df = n-1 = 7-1 = 6. Find the intersection of these row and column values in the table. This is 3.707.)

The head of a computer science department is interested in estimating the proportion of students entering the department who will choose the new computer engineering option. A preliminary sample indicates that the proportion will be around 0.193. What size sample should the department head take if she wants to be 95% confident that the estimate is within 0.04 of the true proportion?

374 (You need to determine the sample size that will ensure the estimate is within the desired margin of error 0.04. The formula is: n = (Z/E)^2 * pi(1-pi) Given the population proportion that is expected and the level of confidence, we have:[(1.96^2)*0.19300*(1 - 0.19300)]/(0.04^2) = 373.958 Now, we always go to the next highest integer to ensure the margin of error is met. So, the answer is 374. Never round down!)

The actual voltages of power packs labeled as 12 volts are as follows: 11.77, 11.90, 11.64, 11.84, 12.13, 11.99, 11.77. A confidence interval for this sample would be based on the t distribution with

6.00 degrees of freedom

After an extensive advertising campaign, the manager of a company wants to estimate the proportion of potential customers that recognize a new product. She samples 120 potential consumers and finds that 54 recognize this product. She uses this sample information to obtain a 95% confidence interval that goes from 0.36 to 0.54. True or False: The parameter of interest to the manager is the proportion of potential customers in this sample that recognize the new product.

F

A confidence interval was used to estimate the proportion of statistics students that are females. A random sample of 72 statistics students generated the following 90% confidence interval: (0.443, 0.637). Based on the interval above, is the population proportion of females equal to 0.60?

Maybe. 0.60 is a believable value of the population proportion based on the information above.

Recall the (1-a)100% confidence interval for the population mean u with known standard deviation o

x bar +/- za/2 *o/sqrtn

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. The 95% confidence interval for p is 0.59 +/- 0.07. Interpret this interval.

We are 95% confident that the true proportion of all students receiving financial aid is between 0.52 and 0.66.

suppose we know the mean exam grades for a sample of 10 students is 90. if you know the first 9 grades, then we also know the tenth, because we know the sum of the ten grades must be 900. therefore, only 9 grades are allowed to freely take on any value. we lose 1 degree of freedom when estimating the sample standard deviation.

df = n-1

When determining the sample size for a proportion for a given level of confidence and sampling error, the closer to 0.50 that p is estimated to be the ---- the sample size required.

larger

Assuming the random variable X is normally distributed, compute the upper and lower limit of the 95% confidence interval for the population mean if a random sample of size n=11 produces a sample mean of 43 and sample standard deviation of 3.68.

lower limit = 40.53 upper limit = 45.57

A random sample of 22 scalpers' ticket prices for a rock concert has sample mean $53.00 and sample standard deviation of $5.68. What is the upper and lower limit of the 95% confidence interval of mean scalpers' ticket prices?

lower limit = 50.48 upper limit = 55.52 (Step 1: Determine the proper tvalue Df = n - 1 = 22 - 1 = 21 The tvalue for a 95% confidence interval and a degrees of freedom of 21 is 2.08. Step 2: Substitute all of the above information into the formula. Lower Limit = 53.00 - 2.08 × (5.68/√22) Lower Limit = 53.00 - 2.5188 = 50.481 = 50.48 Upper Limit = 53.00 + 2.08 × (5.68/√22) Upper Limit = 53.00 + 2.5188 = 55.519 = 55.52

The mean and standard deviation for voltages of power packs labeled as 12 volts for a sample of 25 are as follows: sample mean= 11.43 sample standard deviation= 0.2 Please develop a 90.00% confidence interval for the sample above. (Round to 3 decimal places.)

lower limit: 11.362 upper limit: 11.498

student's t distribution is bell shaped and centered at

mean of 0

An economist is interested in studying the incomes of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What sample size would the economist need to use for a 95% confidence interval if the width of the interval should not be more than $100?

n = 1537

The width of a confidence interval estimate for a proportion will be

narrower for 90% confidence than for 95% confidence.

sample proportion

p = x/n

CLT

p ~ N(pi,op), - op = sqrt of pi(1-pi)/n

CI:

p+/- za/2 * sqrt of p(1-p)/n

population proportion

pi

CI

piE(p-z1-a/2 op, p+z1-1/2 op)

If a decision-maker wishes to reduce the margin of error associated with a confidence interval estimate for a population mean, she can:

reduce the confidence level

we know how to estimate o

s = 1/(n-1) * sum of (xi-xbar)^2

if population standard deviation (op) is unknown, estimate it using

sqrt of p(1-p)/n

replace the z value with the t value from the student's t distribution

xbar +/- ta/2, n-1 s/sqrtn

for any (1-a)100% confidence level and for any sample size,

ta/2 n-1 >= za/2 which implies wider confidence intervals. the smaller the sample size, the larger is the t value for any given level of confidence, and thus the wider the confidence interval.

The margin of error is:

the largest possible sampling error for a specified level of confidence; the critical value times the standard error of the sampling distribution