A researcher is interested in the effect of prenatal alcohol exposure on birth w

Question

A researcher is interested in the effect of prenatal alcohol exposure on birth weight in rats. She hypothesizes that alcohol will have tonic sort of effect, either increasing or decreasing the weight of baby rats. She knows that this species of rat normally produce 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, aid finds the average weight of this sample is x = 14. Slate the null and alternative hypothesis, both in words and in symbol. Determine the Z-critical values that determine the boundaries for the region of rejection (rang alpha = 05) Determine the critical region and Critical values using an alpha level of alpha = .05 for a standard normal distribution (i.e., draw the distribution of z-test scores we would get if the null hypothesis were this. 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.) If the null hypothesis is true, what is the probability that our z-test statistic would fall outside of our z-critical boundaries? Justify your answer. Calculate the standard error (S.F) Calculate your Z-test value, comparing the sample mean to the population mean specified by the null hypothesis Given your Z-test value, what decision should you make regarding the null hypothesis? What conclusion should you make regarding the effort of prenatal alcohol exposure on the birth weight of rats?

Explanation / Answer

Given that,
population mean(u)=18
standard deviation, =6
sample mean, x =14
number (n)=16
null, Ho: =18
alternate,it decreases the weight of body rats 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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