Question

please show work in excel

Question 5 25 pts Answer this question on Recruits sheet. AU.S. Navy recruiting center knows from past experience that the heights of its recruits have traditionally been normally distributed with mean 69 inches. The recruiting center wants to test the claim that the average height of this year's recruits is greater than 69 inches. To do this, recruiting personnel take a random sample of 64 recruits from this year and record their heights. The data are provided in the file Final.xlsx sheet named Recruits. a. Identify the null and alternative hypotheses for this situation b. On the basis of the available sample information, do the recruiters find support for the given claim at the 5% significance level? Explain. For part b, reach conclusion using t-critical value and t-Test Statistic value] c. Find the p-value. HTML Editor BIVA-A-LEE 3 - 2 V B X X, SEE 12pt - Paragraph - 60

Answer 1

5.

Given that,
population mean(u)=69
sample mean, x =70.673
standard deviation, s =2.99
number (n)=64
null, Ho: μ=69
alternate, H1: μ>69
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.669
since our test is right-tailed
reject Ho, if to > 1.669
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =70.673-69/(2.99/sqrt(64))
to =4.4763
| to | =4.4763
critical value
the value of |t α| with n-1 = 63 d.f is 1.669
we got |to| =4.4763 & | t α | =1.669
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 4.4763 ) = 0.00002
hence value of p0.05 > 0.00002,here we reject Ho
ANSWERS
---------------
a.
null, Ho: μ=69
alternate, H1: μ>69
b.
test statistic: 4.4763
critical value: 1.669
decision: reject Ho
c.
p-value: 0.00002
we have enough evidence to support the claim that average height of this year's recruits is greater than 69 inches.