A researcher is interested in the effect of prenatal alcohol exposure on birth w
ID: 3178845 • Letter: A
Question
A researcher is interested in the effect of prenatal alcohol exposure on birth weight in rats. She hypothesizes that alcohol will have some sort of effect, either increasing or decreasing the weight of baby rats. She knows that this species of rat normally produces babies that average a weight of mu = 18 grams, with a population standard deviation of sigma = 6 grams. She then compares the weight of n = 16 rat babies, whose mothers were exposed to alcohol during pregnancy, and finds the average weight of this sample is x = 14. State the null and alternative hypothesis, both in words and in symbols. Determine the Z-critical values that determine the boundaries for the region of rejection (using a = .05). Determine the critical region and Z critical values using an alpha level of a = .05 fora standard normal distribution (i.e., draw the distribution of z-test scores we would get if the null hypothesis were true, shade in the region of rejection and label the probability/proportion for each area (3 of them) under the curve. Don't worry too much about scale.)Explanation / Answer
Given that,
population mean(u)=18
standard deviation, =6
sample mean, x =14
number (n)=16
null, Ho: =18
alternate, H1: <18
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 14-18/(6/sqrt(16)
zo = -2.66667
| zo | = 2.66667
critical value
the value of |z | at los 5% is 1.645
we got |zo| =2.66667 & | z | = 1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : left tail - ha : ( p < -2.66667 ) = 0.00383
hence value of p0.05 > 0.00383, here we reject Ho
ANSWERS
---------------
null, Ho: >18
alternate, H1: <18
test statistic: -2.66667
critical value: -1.645
decision: reject Ho
p-value: 0.00383
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