lOMoARcPSD|4629196
The additional growth of plants in one week are recorded for 11 plants with a sample standard deviation of 2 inches
and sample mean of 12 inches. t* at the 0.05 significance level = Margin of error = Confidence interval =
Chapter 9 Note – Estimation and Confidence Intervals
Point estimate
o A point estimate is a single value (statistic) used to estimate a population value
(parameter)
o Point estimate
The statistic, computed from sample information, that estimates a population
parameter
o Example
Suppose the Bureau of Tourism for Barbados ants to estimate the mean amount
spent by tourists visiting that country. They randomly select 500 tourists as they
depart and ask these tourists about their spending while there. The mean
amount spent by the sample of 500 tourists serves as an estimate of the
unknown population parameter
Confidence intervals
o A confidence interval is a range of values within which the population parameter is
expected to occur
o Confidence interval
A range of values constructed from sample data so that the population
parameter is likely to occur within that range at a specified probability. The
specified probability is called the level of confidence
o The factors that determine the width of a confidence interval for a mean are
The number of observations in the sample, n
The variability in the population, usually estimated by the sample standard
deviation, s
The desired level of confidence
, lOMoARcPSD|4629196
Level of confidence, σ known
o To determine the confidence limits when the population standard deviation is known,
we use the z distribution
x )
x(line bar) – sample mean
z – z-value for a particular confidence level
σ – the population standard deviation
n – the number of observations in the sample
Finding a value of z
o The method for finding z for a 95% confidence interval is:
Divide the confidence interval in half
. = .4750
Find the value .4750 in the body of the table
Identify the row and colum and add the value
The probability of finding a value between 0 and 1.96 is .4750
So the probability of finding a value between +/- 1.96 is .9500
The additional growth of plants in one week are recorded for 11 plants with a sample standard deviation of 2 inches
and sample mean of 12 inches. t* at the 0.05 significance level = Margin of error = Confidence interval =
Chapter 9 Note – Estimation and Confidence Intervals
Point estimate
o A point estimate is a single value (statistic) used to estimate a population value
(parameter)
o Point estimate
The statistic, computed from sample information, that estimates a population
parameter
o Example
Suppose the Bureau of Tourism for Barbados ants to estimate the mean amount
spent by tourists visiting that country. They randomly select 500 tourists as they
depart and ask these tourists about their spending while there. The mean
amount spent by the sample of 500 tourists serves as an estimate of the
unknown population parameter
Confidence intervals
o A confidence interval is a range of values within which the population parameter is
expected to occur
o Confidence interval
A range of values constructed from sample data so that the population
parameter is likely to occur within that range at a specified probability. The
specified probability is called the level of confidence
o The factors that determine the width of a confidence interval for a mean are
The number of observations in the sample, n
The variability in the population, usually estimated by the sample standard
deviation, s
The desired level of confidence
, lOMoARcPSD|4629196
Level of confidence, σ known
o To determine the confidence limits when the population standard deviation is known,
we use the z distribution
x )
x(line bar) – sample mean
z – z-value for a particular confidence level
σ – the population standard deviation
n – the number of observations in the sample
Finding a value of z
o The method for finding z for a 95% confidence interval is:
Divide the confidence interval in half
. = .4750
Find the value .4750 in the body of the table
Identify the row and colum and add the value
The probability of finding a value between 0 and 1.96 is .4750
So the probability of finding a value between +/- 1.96 is .9500
