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WorksheetsChapter 8 for AP Stats
Total questions: 20
Worksheet time: 16mins
A Confidence Interval consists of:
statistic
t and z
statistic, critical value, and standard deviation of statistic
statistic, critical value, and probability
The correct formula for a Confidence Interval is:
Statistic ± (critical value) * (standard deviation of statistic)
Statistic + (critical value) * (standard deviation of statistic)
Statistic + (critical value) * (confidence level)
Critical Value ± (statistic) * (sample size)
To interpret a Confidence Level you would say:
95% of all possible samples of a given size from a population will result in a probability greater than the alpha level
95% of all samples of a given size will result in an interval that is greater than the unknown parameter
95% of all samples of a given size will result in an interval that is smaller than the unknown parameter
95% of all possible samples of a given size from a population will result in an interval that captures the unknown parameter
What is the correct way to interpret a Confidence Interval?
The interval from .375 to .478 does not capture the parameter value.
We are 95% confident that the interval from .375 to .478 captures the actual value of the population parameter
We are 95% confident that the interval from .375 to .478 captures the actual value of the confidence level
95% of all possible samples of a given size from a population will result in a interval that capture the unknown parameter
Confidence Intervals are statements about:
parameters
statistics
confidence levels
chi square statistics
If you were given
statistic: .426
Critical Value: 1.96
Standard Deviation: .031
the correct way to construct an interval would be:
.426 + (1.96) * (.031)
.426 ± (1.96) * (.031)
1.96 ± (.426) * (.031)
.426 ± (1.96)
The statistic for a 95% confidence level would be
1.96
1.28
2.575
2.326
What would the calculator function be when finding the critical value of a 99% Confidence Level?
InvNorm(.99, 0, 1)
InvNorm(.90, 0, 1)
InvNorm(.99, 1, 0)
InvNorm(.995, 0, 1)
What is the Independent condition?
Condition stating that the individual observations are independent and that our sample size is less than a tenth of the population.
Condition stating that our successes and failures must be greater than 10
Condition stating that the individual observations are independent
Condition stating that our successes and failures must be less than 10
What are the appropriate conditions that must be met first in order to estimate a confidence interval?
The successes and failures are greater than 10
Random, Normal, Independent
Random
Random and 10% condition
Which condition is different for proportions when comparing it to the conditions that must be met for means?
Normal
Random
Confidence Level
Random and Confidence Level
What is the correct formula for a One Sample Z Interval for Population Proportions?
( z*) √( (p̂(1- p̂) / n)
p̂ ± ( z*) √( n)
p̂ ± ( z*) √( (p̂(1- p̂) )
p̂ ± ( z*) √( (p̂(1- p̂) / n)
When using the 4 step process, what must you always include in the beginning of the DO part?
"conditions have been met"
the confidence level
the confidence interval
"If conditions are met"
Which formula will correctly give you the sample size needed for an interval involving proportions?
√( (p̂ (1-p̂) / n ) ≤ ME
t* √( (p̂ (1-p̂) / n ) ≤ ME
z* √( (p̂ (1-p̂) / n ) ≤ ME
z* √( (p̂ (1-p̂) / n ) ≥ ME
When given the information below for proportions:
z* = 1.96
p̂ = .5
ME= .03
the sample size would be?
100
1068
1060
10068
What is the correct formula to find a one sample t interval?
p̂
x̅ ± (t*) ( p̂/ √(n) )
z ± (t*) ( Sx/ √(n) )
x̅ ± (t*) ( Sx/ √(n) )
What would you use to find a sample size needed to obtain a confidence interval with margin of error ME for population means
z* ( σ/ (√n) ) ≤ ME
( σ/ (√n) ) ≤ ME
σ ≤ ME
Confidence level
What would be the degrees of freedom when using a sample size of 50 in a t distribution?
51
50
49
47
When constructing a confidence interval for sample means, the correct t* for a confidence level of 98% and a sample size of 22 would be?
InvT(.99, 21) = 2.518
InvNorm(.98, 0, 1) = 2.053
InvT(.98, 21) = 2.189
InvNorm(.99, 0, 1) = 2.326
If you aren't told that a populations distribution is Normal then what 2 other methods should you check for in order to satisfy the Normal condition?
random and independence
np ≥ 10 and n(1-p) ≥ 10
check that it is random
CLT n ≥ 30 or sketch a graph and check for outliers or strong skewness
