Please answer all questions if you are going to answer. Thank you Problem 2 (12

Question

Please answer all questions if you are going to answer. Thank you

Problem 2 (12 points) A company that manufactures hearing aids wants to test to see ifa new hearing aid helps senior citizens hear more words. Six senior citizens who currently wear hearing aids are recruited. They listen to a tape with 50 words on it and repeat each word they hear back to the tester who keeps track of the number of correct words. After an hour long break, the subjects put in the new hearing aid and repeat the process with a different set of 50 words. The tapes are believed to be of equal difficulties. Assume the differences in the word counts are normally distributed. Use the data in Minitab to determine if the new hearing aid helps senior citizens hear significantly more words than their old hearing aid using a 1% level of significance. a. What type of test should be used here? Briefly explain why. (2 points) Type of Test: Reason b. Write the null and alternative hypotheses. HO: c Check the conditions to ensure that the test from part (a) is appropriate to use. d. Calculate the test statistic. (You may use Minitab and just report the value, but be sure you can do the calculation by hand as well.)

Explanation / Answer

Given that,
population mean(u)=30
sample mean, x =27
standard deviation, s =22
number (n)=32
null, H0: Ud > 0
alternate,significantly hear more words than their old hearing aid H1: Ud < 0
level of significance, = 0.01
from standard normal table,left tailed t /2 =3.365
since our test is left-tailed
reject Ho, if to < -3.365
we use Test Statistic
to= d/ (S/n)
Where
Value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -9.17
We have d = -9.17
Pooled variance = Calculate value of Sd= S^2 = Sqrt [ 695-(-55^2/6 ] / 5 = 6.18
to = d/ (S/n) = -3.64
Critical Value
The Value of |t | with n-1 = 5 d.f is 3.365
We got |t o| = 3.64 & |t | =3.365
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
p-value :left tail - Ha : ( p < -3.6358 ) = 0.00748
hence value of p0.01 > 0.00748,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud > 0
alternate, H1: Ud < 0
test statistic: -3.64
critical value: reject Ho, if to < -3.365
decision: Reject Ho
p-value: 0.00748

X Y X-Y (X-Y)^2 42 43 -1 1 33 41 -8 64 30 40 -10 100 27 39 -12 144 22 41 -19 361 32 37 -5 25 -55 695

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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