11. A simple random sample of 81 births of Chinese babies resulted in a mean bir
ID: 3324555 • Letter: 1
Question
11. A simple random sample of 81 births of Chinese babies resulted in a mean birth weight of 3245 g and a standard deviation of 466 g. Use a significance level of 0.01 and follow the steps for the -Value Method below to test the claim that the mean birth weight of Chinese babies is less than 3369 g (which is the mean birth weight for Caucasian babies). a) State the null and alternative hypotheses, and identify the claim. Ho: b) Circle the distribution you will use: Standard Normal Distribution t-Distribution 2-Distribution (2 pts) Circle the test type Left-tailed Right-tailed Two-tailed 2 pts) c) Calculate the test statistic and find the P-value for this hypothesis test. Round the test statistic to 3 decimal places and the Pivalue to 3 significant digits (4 pts) Use your answers to parts (b) and (c), to make a decision. Circle the correct decision: d) 2 pes) Reject Ho Fail to Reject Ho e) Write your conclusion using regular words that address the original claim. (2 pts)Explanation / Answer
Given that,
population mean(u)=3369
sample mean, x =3245
standard deviation, s =466
number (n)=81
null, Ho: =3369
alternate, H1: <3369
level of significance, = 0.01
from standard normal table,left tailed t /2 =2.374
since our test is left-tailed
reject Ho, if to < -2.374
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =3245-3369/(466/sqrt(81))
to =-2.3948
| to | =2.3948
critical value
the value of |t | with n-1 = 80 d.f is 2.374
we got |to| =2.3948 & | t | =2.374
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -2.3948 ) = 0.00948
hence value of p0.01 > 0.00948,here we reject Ho
ANSWERS
---------------
t distribution
null, Ho: =3369
alternate, H1: <3369
left tailed test
test statistic: -2.3948
critical value: -2.374
decision: reject Ho
p-value: 0.00948
have evidence that chinese babies is less than 3369
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