Answer:
The 90% confidence interval is (31.78 grams, 33.62 grams).
Explanation:
The first step is finding our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = (1-0.9)/(2) = 0.05](https://img.qammunity.org/2020/formulas/mathematics/college/z1qgp6bnfq57huolcnb6k5emwlwxzsxx5l.png)
Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, that is between
and
, so we use
.
Now, find M as such:
![M = z*(\sigma)/(√(n)) = 1.645*(5)/(√(80)) = 0.92](https://img.qammunity.org/2020/formulas/mathematics/college/4bkwswv310bd2dbncx2dk7iix0xm1d1873.png)
In which
is the standard deviation and n is the length of the sample
The lower end of the interval is the sample mean subtracted by M. So it is 32.7 - 0.92 = 31.78 grams.
The upper end of the interval is the sample mean added to M. So it is 32.7 + 0.92 = 33.62 grams.
The 90% confidence interval is (31.78 grams, 33.62 grams).