Dallas Tampa Bay n = 22 n = 45 xbar = 59700 xbar = 55900 s = 6700 s = 5700 Quest

Question

Dallas Tampa Bay
n = 22 n = 45
xbar = 59700 xbar = 55900
s = 6700 s = 5700
Question 2A Based on the data, can you say the average nursing salary in Tampa is smaller than the average nursing salary in Dallas at =0.025?
State the hypothesis in terms of Tampa-Dallas.
Step 1: H0:
Ha:
Step 2: =
Step 3: Fill in row 88 if the hypothesis is one tailed.
Reject H0 if
Fill in row 92 if the hypothesis is two tailed. The smaller number must be typed first.
Reject H0 if or
Step 4: =
Step 5: H0
Step 6:
Step 7: Fill in row 102 if the Z table is appropriate.
P-value =
Fill in row 106 if the t table is appropriate. The lower boundary must go on the left and the upper boundary on the right.
P-value
Question 2B Construct a 98% confidence interval for the difference in true means. State the C.I. in terms of Tampa-Dallas.
Lower Endpoint Upper Endpoint

Explanation / Answer

Given that,
mean(x)=59700
standard deviation , s.d1=6700
number(n1)=22
y(mean)=55900
standard deviation, s.d2 =5700
number(n2)=45
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.025
from standard normal table,right tailed t /2 =2.08
since our test is right-tailed
reject Ho, if to > 2.08
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =59700-55900/sqrt((44890000/22)+(32490000/45))
to =2.2863
| to | =2.2863
critical value
the value of |t | with min (n1-1, n2-1) i.e 21 d.f is 2.08
we got |to| = 2.28631 & | t | = 2.08
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 2.2863 ) = 0.01636
hence value of p0.025 > 0.01636,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 2.2863
critical value: 2.08
decision: reject Ho
p-value: 0.01636

we have evidence to claim that nursing salary in Tampa is smaller than the average nursing salary in Dallas

b.
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 59700-55900) ± t a/2 * sqrt((44890000/22)+(32490000/45)]
= [ (3800) ± t a/2 * 1662.063]
= [-385.075 , 7985.075]

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

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