Assume that a simple random sample has been selected from a normally distributed

Question

Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative hypotheses, test statistic, critical value(s), and state the final conclusion that addresses the original claim. A simple random sample of pages from a dictionary is obtained. Listed below are the numbers of words defined on those pages. Given that this dictionary has 1459 pages with defined words, the claim that there are more than 70,000 defined words is the same as the claim that the mean number of defined words on a page is greater than 48.0. Use a 0.10 level significance level to test this claim. What does the result suggest about the claim that there are more than 70,000 defined words in the dictionary? 62 49 53 64 48 71 44 47 121 71 What are the null and alternative hypotheses? A. H_0: mu = 48.0 H_1: mu 48.0 D. H_0: mu > 48.0 H_1: mu = 48.0 Identify the test statistic.

Explanation / Answer

Given that,
population mean(u)=48
sample mean, x =63
standard deviation, s =22.6863
number (n)=10
null, Ho: <=48
alternate, H1: >48
level of significance, = 0.1
from standard normal table,right tailed t /2 =1.383
since our test is right-tailed
reject Ho, if to > 1.383
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =63-48/(22.6863/sqrt(10))
to =2.091
| to | =2.091
critical value
the value of |t | with n-1 = 9 d.f is 1.383
we got |to| =2.091 & | t | =1.383
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 2.0909 ) = 0.03304
hence value of p0.1 > 0.03304,here we reject Ho
ANSWERS
---------------
null, Ho: <=48
alternate, H1: >48
test statistic: 2.091
critical value: 1.383
decision: reject Ho
p-value: 0.03304

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.