9.
The conditions are present to construct a one-proportion z interval.
We are 95% confident that the true proportion of people who will get the flu despite getting the
vaccine is between 11.9% and 19.5%. To say that we are 95% confident means that the process used
to construct this interval will, in fact, capture the true population proportion 95% of the time (that is,
our “confidence” is in the procedure, not in the interval we found!).
(a) We are 95% confident that the true proportion of subjects helped by a new anti-inflammatory
drug is (0.56, 0.65).
(b) The process used to construct this interval will capture the true population proportion, on
average, 95 times out of 100.
- We have . Because the hypothesis is two sided, we are
concerned about the probability of being in either tail of the curve even though the finding was larger
than expected.
Then, the P -value = 2(0.1788) = 0.3576. Using the TI-83/84, the P -value =
2normalcdf(0.92,100) .
The problem states that the samples were randomly selected and that the scores were approximately
normally distributed, so we can construct a two-sample t interval using the conservative approach.
For C = 0.99, we have df = min{23 – 1, 26 – 2} = 22 t = 2.819.
Using a TI-84 with the invT function, t = invT(0.995,22 ) .
0 is a likely value for the true difference between the means because it is in the interval. Hence, we
do not have evidence that there is a difference between the true means for males and females.
Using the STAT TESTS 2-SampTInt function on the TI-83/84, we get an interval of (-5.04, 7.24), df
= 44.968. Remember that the larger number of degrees of freedom used by the calculator results in a
somewhat narrower interval than the conservative method (df = min{n 1 – 1, n 2 – 1}) for computing
degrees of freedom.
The power of the test is our ability to reject the null hypothesis. In this case, we reject the null if >
36.5 when μ = 38. We are given s = 0.99. Thus