b. Idenfity the critical value(s) ______(Round to three decimal places as needed
ID: 3326636 • Letter: B
Question
b. Idenfity the critical value(s)
______(Round to three decimal places as needed.)
Score: 0.5 of 1 pt 25 of 35 (30 complete) HW Score: 71.89%, 25.16 of 35 pts &8.4.23 Question Help * 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 0.05 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? 61 48 66 80 56 83 59 53 97 77 Click the icon to view a table of critical t-values. What are the null and alternative hypotheses? A. Ho: = 48.0 H1 : #48.0 Ho: =48.0 H1 : > 48.0 OB. HOP>48.0 H1:=48.0 D. Ho:H 48.0 H1:Explanation / Answer
Given that,
population mean(u)=48
sample mean, x =68
standard deviation, s =15.6134
number (n)=10
null, Ho: =48
alternate, H1: >48
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.833
since our test is right-tailed
reject Ho, if to > 1.833
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =68-48/(15.6134/sqrt(10))
to =4.0507
| to | =4.0507
critical value
the value of |t | with n-1 = 9 d.f is 1.833
we got |to| =4.0507 & | t | =1.833
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 4.0507 ) = 0.00144
hence value of p0.05 > 0.00144,here we reject Ho
ANSWERS
---------------
null, Ho: =48
alternate, H1: >48
test statistic: 4.051
critical value: 1.833
decision: reject Ho
p-value: 0.00144
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