For the following claim, find the null and alternative hypotheses, test statisti
ID: 3224213 • Letter: F
Question
For the following claim, find the null and alternative hypotheses, test statistic, critical value, and draw a conclusion. Assume that a simple random sample has been selected from a normally distributed population. Answer parts a-d. Claim: The mean IQ score of statistics professors is less than 116. Sample data: n=18, x=112, s=4. The significance level is s=0.05. a. Choose the correct null hypothesis (Upper H 0H0) and alternative hypothesis (Upper H 1H1). A. Upper H 0H0: muless than<116116 Upper H 1H1: mugreater than>116116 B. Upper H 0H0: muequals=116116 Upper H 1H1: munot equals116116 C. Upper H 0H0: muequals=116116 Upper H 1H1: muless than<116116 D. Upper H 0H0: muless than<116116 Upper H 1H1: muequals=116
Explanation / Answer
Given that,
population mean(u)=116
sample mean, x =112
standard deviation, s =4
number (n)=18
null, Ho: =116
alternate, H1: <116
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.74
since our test is left-tailed
reject Ho, if to < -1.74
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =112-116/(4/sqrt(18))
to =-4.243
| to | =4.243
critical value
the value of |t | with n-1 = 17 d.f is 1.74
we got |to| =4.243 & | t | =1.74
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -4.2426 ) = 0.00027
hence value of p0.05 > 0.00027,here we reject Ho
ANSWERS
---------------
null, Ho: =116
alternate, H1: <116
test statistic: -4.243
critical value: -1.74
decision: reject Ho
p-value: 0.00027