(2 points) A study is conducted to determine if a newly designed text book is mo
ID: 3356891 • Letter: #
Question
(2 points) A study is conducted to determine if a newly designed text book is more helpful to learning the material than the old edition. The mean score on the final exam for a course using the old edition is 75. Scores follow a normal distribution. Ten randomly selected people who used the new text take the final exam. Their scores are shown in the table below Person ABCDEF GHI Test Score 6976709396 838790 8574 Use a 0.01 significance level to test the claim that people do better with the new edition. (1.) What kind of test should be used? (0) A. One-Tailed O B. Two-Tailed O C. It does not matter. (rounded to 2 decimals). (2.) The test statistic is 2.54 (3.) The P-value is 032 (4.) Is there sufficient evidence to support the claim that people do better than 75 on this exam? A. No OB, Yes (5) Construct a 99% confidence interval for the mean score for students using the new text.Explanation / Answer
Given that,
population mean(u)=75
sample mean, x =82.3
standard deviation, s =9.5922
number (n)=10
null, Ho: = 75
alternate, H1: >75
level of significance, = 0.01
from standard normal table,right tailed t /2 =2.8214
since our test is right-tailed
reject Ho, if to > 2.8214
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =82.3-75/(9.5922/sqrt(10))
to =2.407
| to | =2.407
critical value
the value of |t | with n-1 = 9 d.f is 2.8214
we got |to| =2.407 & | t | =2.8214
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > 2.4066 ) = 0.01973
hence value of p0.01 < 0.01973,here we do not reject Ho
ANSWERS
---------------
a. one tailed
b. test statistic = 2.54
c. p-value: 0.01973
d. yes suffcient evience
e.
confidence interval = [ 82.3 ± t a/2 ( 9.5922/ Sqrt ( 10) ]
= [ 82.3-(3.25 * 3.033) , 82.3+(3.25 * 3.033) ]
= [ 72.442 , 92.158 ]
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.