Statistics final exam from professor
To find probabilities for sums on the calculator, follow these steps.
2nd DISTR 2:normalcdf The parameter list is abbreviated normalcdf (lower value of the area, upper value of the area, (n)(mean), (n√n)(standard deviation)
Sampling Distribution
This distribution of sample means The probability distribution of a sample statistic when a sample is drawn from a population. A probability distribution consisting of all possible values of a sample statistic.
invNorm
find proportions or probabilities of sampling distribution
σ/√η
formula for sample standard deviation in a sampling distribution
n= area of z²x pˆx qˆ /mee²
how do you find How many families will you have to survey
p hat
sample proportion
mean of the sample sums
will equal n times the population mean
Margin of error
Critical value x Standard deviation of the statistic Critical value x Standard error of the statistic
97 (z = 1.96, m.e.e = $1,000 and & M $5,000 into formula - n=(z²)(q²)/mee²
A social worker is given the task of finding a 95% confidence interval on the mean income of families in Chatham county who receive federal aid. The width of the interval cannot exceed $2000. The estimated standard deviation of these incomes is $5,000. How large of a sample is needed?
σ
Standard deviation of population mean Sigma (sometimes has underscore x without line over) - from previously collected data
pˆ = qˆ = 0.5.
When no information about the proportion are available,
2401(MEE= .02 ( interval is .04 MEE is 1/2 or each side of mean) CI =.95 alpha = 1 - 0.95 = 0.05 critical standard score (z) 1 - alpha/2 = 1 - 0.05/2 = 0.975. using invnorm z=1.96. p and q= .5 n = 1.96²x.5x.5/.02²)
You are given the task of finding a 95% confidence interval on the population proportion of families in the greater Savannah area who have at least one child at home. You are required to have an interval no wider than 0.04. How many families will you have to survey?
η
represents the number of data entries given. (Central Limit Theorem)- 20 people were asked
X
sample mean (symbol x) means the data used is part of a large group. is a random variable with a distribution that may be known or unknown
qˆ
sample proportion 1-pˆ equals what
s
standard deviation for sample data is the standard deviation of sample means, or standard error of the mean.
Z distribution (normcdf)
standard deviation of the population is known
Student's t-distribution
standard deviation of the sample size is known
(n√)(σX)
standard deviation of ΣX
critical value
t score or as a z score. Z score when population standard deviation is known
Standard Error of the Mean
the standard deviation of the distribution of the sample means. formula = σ/√η
σ (sub x-bar)
the symbol for standard deviation of the distribution of the sample means or standard error of the mean. ( sample data collected) Formula for finding = σ/√η(standard error)
mean = 518 (Mean of population approximates sample mean) 17.16 (σS/SD of these sample means =σ/√η=94/√30=17.16) Normal (sampling size is over 30)
, population mean 518 population standard deviation 94 number sampled 30 What the mean of the samples means? What is the SD of these samples? What shape will this sampling distribution have?
As the sample size of the sample means increases
, the mean of the sampling distribution will remain the same and the standard deviation will decrease resulting in a skinnier & taller sampling distribution.
k (percentile or proportion)
invNorm (area to the left of k, (n)(mean), (n√)(n)(standard deviation) k is the kth percentile mean is the mean of the original distribution standard deviation is the standard deviation of the original distribution sample size = n
(n)(µ×)
The normal distribution has a mean equal to the original mean multiplied by the sample size
A) T interval (pop SD is known) (98.096 , 98.402) B) no (the value 98.6 is not in the interval)
6. A sample of 130 healthy adults participated in a health screening. From this sample the mean body temperature was determined to be 98.25˚ F. Assume the population standard deviation in known to be 0.75˚ F. A) Find a 98% confidence interval for the mean body temperature of healthy adults. B) Using the interval from part A of this question, can we claim with 98% confidence that the mean body temperature of healthy adults is 98.6˚ F?
Student's t-distribution
T ~ t underscore df where (df = n - 1)
234 (n= amount of debt z= 1.96 (area of z with 95% CI) σ(bar) = 780 MEE=100 n=(zxσ/mee)² (formula used when population size is unknown and the sample statistic is known. ))
A researcher wants to estimate the amount of debt students collect per semester while attending a state university. It is desired that the M.E.E. be no larger than $100 in a 95% confidence interval. Assume the standard deviation is known to be $780.
proportion, mean, proportion, mean
A survey given to adults in Chatham county contained the following questions. a) Was your gross household income more than $40,000 last year? b) How many dependents could you claim on your taxes last year? c) Do you have a mortgage on your place of residence? d) How much personal debt do you currently have? For each of these four questions, determine if it would be more appropriate to calculate a confidence interval for a proportion or a confidence interval for a mean using the data
(1-PropZInt) .14237,.19763 (We are 90% confident the population proportion of Georgians who have absolutely no worries about the economy going into a slump is between 0.142 and 0.198.)
A survey of 500 people from Georgia revealed that 85 have absolutely no worries about the economy going into a slump. Find a 90% confidence interval for the population proportion of Georgians who have absolutely no worries about the economy going into a slump.
0.046 to 0.110 ( use PropZint) No ( entire interval is not below .08)
A survey was recently conducted on college campuses in Georgia inquiring about credit card debt. From the 385 surveys collected, it was determined that 7.8% of students have more than $500 of credit card debt. A) Find a 98% confidence interval for the proportion of college students in Georgia with more than $500 of credit card debt. B) Using your confidence interval, can you confidently say that less than 8% of college students have more than $500 of credit card debt? Why or why not?
361.32,378.69 9 ( high end -mean) no - mean not individual students
After recently purchasing your textbooks for the semester, you wonder, on average, how much does a typical full time AASU student pay for textbooks each semester. You decide to estimate this population mean by finding a 90% confidence interval on the mean amount of money a full time AASU student pays for textbooks per semester. From a random sample of 41 full time students, you calculate a mean of $370 and a standard deviation of $33. A) Find the 90% confidence interval. B) What is the maximum error of the estimate for the point estimate $370 at a 90% level of confidence? C) Can you claim that 90% of full time students pay more than $360? 15.
T interval ( SD is from sample) 2.76,3.19 yes because the entire interval is higher than 2.50
As a student you wonder what the mean GPA is for all students in the College of Science and Technology at AASU. From a random sample of 30 AASU students in the College of Science and Technology you calculate a mean GPA of 2.98 with a standard deviation of 0.55. A) Find a 96% confidence interval for the population mean GPA of all students in the College of Science and Technology. B) Using your confidence interval, can you claim with 96% confidence that the mean GPA of students in the College of Science and Technology is greater than 2.50? Why or why not?
(1-PropZInt) A) .5125 or 51.25 % B) 0.4842, 0.5408 (We are 95% confident the population proportion of American citizens who consider themselves as conservative is between 0.4842 and 0.5408) C) No ( there are values in the interval that are less than 0.50 (50%))
From a random sample of 1200 American citizens , 615 declare themselves as conservative. A) Find the sample proportion of American citizens who consider themselves as conservative (i.e., the point estimate). B) Find a 95% confidence interval for the population proportion of American citizens who consider themselves as conservative. C) According to part B of this question, can we claim, with 95% confidence, that a majority of American citizens consider themselves as conservative?
maximum error of the point estimate
High CI - point estimate of mean
The Central Limit Theorem for Sample Means
In a population whose distribution may be known or unknown, if the size (n) of samples is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size (n).
point estimate
In simple terms, any statistic can be a . A statistic is an estimator of some parameter in a population. For example: The sample standard deviation (s) is a point estimate of the population standard deviation (σ). The sample mean (̄x) is a point estimate of the population mean, μ
54,430 4720.64871 right .988102945 .3476406287 A) $54,430 B) $4720.64 C) normal D) i) z = -.94 (Z-test) => normalcdf((-10. -0.94) = 0.1736 ii) z = 1.18 (z test) => normalcdf(1.18, 10) = 1 - .8810 = 0.119
In the city of Everest, the mean household income is $54,430 and the standard deviation of household incomes is $33,380. A random sample of 50 households will be selected and the mean household income determined. This sample mean is one of the values in a sampling distribution. A) What do we expect the mean of the sampling distribution to be? B) What do we expect the standard deviation of the sampling distribution to be? C) The distribution of household incomes is definitely skewed to the right. What will the shape of the sampling distribution be for random samples of size 50? D) What is the probability the mean income of these 50 households will be i) at most $50,000? ii) at least $60,000?
increase sample size, decrease level of confidence
List two ways to reduce the width of a confidence interval.
use Tinterval data (no sd ) 8.85, 13.15
Safety concerns grow as jets in the U.S. commercial fleet become older. The ages (in years) of 26 randomly selected Boeing 737 jets are listed below: 12 15 7 4 13 7 11 12 13 2 19 7 12 22 18 1 11 17 11 7 9 15 13 6 12 10 Create a 96% confidence interval for the true population mean age of 737 jets in the U.S. commercial fleet.
(t-Interval option) 135.4 116.0, 154.8 19.4 (154.8 - 135.4)
The following 10 values were obtained by checking the LDL cholesterol levels of adult males. Assume these values are from an approximately normal distribution. 135, 165, 122, 109, 98, 190, 176, 140, 88, 131A) Find a point estimate of the population mean LDL cholesterol level of adult males. B) Find a 90% confidence interval for the population mean LDL cholesterol level of adult males. C) Find the of your answer in part A at a 90% level of confidence.
Central Limit Theorem
The larger the sample, the closer the sample mean is to the population mean. When population mean and standard deviation are given
(σ×)(√n)
The normal distribution has standard deviation equal to the original standard deviation multiplied by the square root of the sample size.
A) 518, 17.16, bell B)
The population mean SAT Math score for 2006 was 518 with a standard deviation of 94. Several random samples of thirty students who took the SAT were selected and the average SAT Math scores were determined. A) What should the mean of these sample means be? What should the standard deviation of these sample means be? What shape will this sampling distribution have? B) One sample of 30 students who took the SAT will be randomly selected and the mean SAT Math score determined. What is the probability the sample mean SAT Math score is higher than 530?
normalcdf (normalcdf(lower value of the area, upper value of the area, mean, standard deviation/√sample size) where: mean is the mean of the original distribution standard deviation is the standard deviation of the original distribution sample size = n)
To find probabilities for means on the calculator, follow these steps.
1.96²x.5x.5/.05²= 384 (383.19 rounded up) 1.96²x.3x.7/.05²=323
You are given the task of finding a 95% confidence interval on the population proportion of families in the greater Savannah area who have at least one child in private school. You are required to have a maximum error or the estimate no larger than 0.05. A) How many families will you have to survey? (be conservative in your response) B) If a good guess at the proportion of families with at least on child in private school is 0.30, how many fewer families will you have to survey than in the previous question?
(n)(μX)
mean of ΣX
ΣΧ (sum X)
means the sum of all the x values. Σx is one sum. ΣX consisting of sums tends to be normally distributed and ~ N((n)(μΧ), (n√n)(σΧ)).
ensures a normally shaped sampling distribution of sample means,
minimum sample size of about 30, is needed. However, we do not need this large of a sample if the original population (i.e. the parent population) is approximately normally distributed
µ (symbol μ)
population mean: means the data includes everybody. population mean (Central Limit Theorem) data already collected - assumed
µ (sub x-bar)
symbol for the population mean of sampling distribution (collected data you are comparing ) sample mean is equal/estimated to population mean (Central Limit Theorem)
t= xbar-µ s/√n (sample mean-pop mean/sample SD/√N)
t score formula
degrees of freedom (df)
the degrees of freedom are equal to the number of observations minus one.
if the size (n) of the sample is sufficiently large
the distribution of the sample means and the distribution of the sample sums will approximate a normal distributions regardless of the shape of the population. The mean of the sample means will equal the population mean, and . The standard deviation of the distribution of the sample means, σn√σn, is called the standard error of the mean.
find z score
z = (X - μ) / σ
Find Σx with z-score
Σx = (n)(μX) + (z)(n√)(n)(σΧ)
Compute alpha (α)
α = 1 - (confidence level / 100)