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Notes 9.3
Name____________________
Confidence Intervals for Population Mean
1. A random sample of 100 American men yielded a sample mean of 70 in. and a standard deviation of 6.3 in.
s
Construct 90% confidence interval for  =the true mean height of American men.
x  T /2 
n
Step 1: Find the Critical Value (T score).
_________
To find the T score: Use back cover of text book, or invT ( / 2, df ) . Degrees of Freedom = n-1
Important properties of t distribution:
1. The t distribution corresponding to any fixed number of degrees of freedom is bell shaped and centered at
zero. (Just like the z dist)
2. Each t distribution is more spread out than the standard normal (z) distribution.
3. As the number of degrees of freedom increases, the spread of the corresponding t distribution decreases.
4. As the number of degrees of freedom increases, the corresponding sequence of t distributions approaches
the standard normal (z) distribution.
Step 2: Find the sd.
s
n
Step 3: Use the formula.
Requirements:
_________
x  T  sd
Math
(___ , ___)
Conclusion: _______________________________________________________________________
________________________________________________________________________
2. Estimate the true average test scores taken from a sample of 41 scores. The sample has a x =710 and s=25
Step 1: Find the Critical Value (T score).
s
Step 2: Find the sd.
n
Step 3: Use the formula.
x  T  sd
Requirements:
_________
_________
Math
(___ , ___)
Conclusion: _______________________________________________________________________
________________________________________________________________________
3. The two intervals (92.268 , 97.732) and (90.222 , 99.778 ) are confidence intervals for  = the true average
resonance frequency (in hertz) for all tennis rackets of a certain type.
a. What is the value of the sample mean resonance frequency?
_________
b. What is the margin of error? For both intervals.
____ , _____
c. The confidence level for one of these intervals is 90% and for the other is 99%. Which is which and how
can you tell?
4. Interpret the meaning of 95% confidence. ____________________________________________________
________________________________________________________________________________________
5. A survey of 1000 randomly selected Canadians about how much they will spend on Halloween resulted in
a sample mean of $46.65 and a standard deviation of $83.70.
a. Explain how it could be possible for the standard deviation of the anticipated Halloween expense to be
larger than the mean anticipated expense.
b. Is it reasonable to think that the distribution of the variable anticipated Halloween expense is approximately
normal? Why or why not?
c. Is it appropriate to use the t confidence interval to estimate the mean anticipated Halloween expense for
Canadian residents? Explain why or why not?
d. If appropriate, construct and interpret a 99% confidence interval for the mean anticipated Halloween
expense for Canadian residents.
Note:
You can use Z interval when Population standard deviation(  ) is known. This doesn’t happen very often in
practice. We generally don’t use it. (but sometimes you will see this in AP practice questions) Our rule is:
p̂ Problems: use Z
x Problems: use t