Decision Reject H0
p-value
0.00
the null is rejected.
greater than 16.
n*pi>5
1
n*(1-pi)>5
1
0.975 adj. for ND
-1.96
Right Tailed test
CV
1.64
Decision Do not reject H0
p value
0.155
the null is not rejected
on is not less than 0.63.

it follows a normal decision
0.95 adj. for ND
-1.64
Right-tailed Test
CV
1.28
Decision Do not reject H0
p-value
0.50
, the null is not rejected.
not different from 10 lbs.
9
7.6
7.3
1.2
6
8.8
6.8
8.4
6.6
it follows a normal decision
0.75 adj. for ND

-0.67
Right-tailed Test
CV
0.00
Decision Reject H0
p-value
0.02
e null is not rejected.
than 6.6. lb.
Women
Men
Label
25
30
mean
14
11
SD
30
25
Sample size
est)
Women
25 mean
14 std. dev.
30 n
5.000 difference (Men - Women)
3.372 standard error of difference
0 hypothesized difference
1.48 z
.1382 p-value (two-tailed)
a (0.1), the null is not rejected.
n is greater than pop. Of men.

rtions
p2
0.66
0.6657
p (as decimal)
66/100
233/350
p (as fraction)
66
233
X
100
350
n
0.008 difference
0 hypothesized difference
0.0558 std. error
0.14 z
.8860 p-value (two-tailed)
.5), the null is not rejected.
the pop proportion of adults and children.
it follows a normal decision
0.99 adj. for ND
-2.33
Right-tailed Test
CV
2.05
Decision Do not reject H0
p-value
0.42
p
c

p-value
Err:502
null is not rejected
not less than 60,000.
it follows a normal decision
0.99 adj. for ND
-2.33
Right-tailed Test
CV
2.05
Decision Do not reject H0
p-value
0.04
p-value
Err:502
s not rejected
not less than 60,000.

Mean
SD

Mean
SD

Assignment 20 // Pg 366 #'s 46, 48, 50, 52
Question # 46
T-Test One Sample test of hypothesis pop mean, pop SD unknown
321
Step 1
H0: μ ≤ 267
286
290
330
Given Data
310
ypothesized (H0) pop mean mu
267
250
sample mean
288.31
270
sample size
13
280
sample SD
22.4588
299
265
sig level alpha
0.01
291
275
degrees of freedom
12
281
Calculated Test-Statistic
T score
3.42
Two tailed test
CV
3.0545
-3.0545
Decision Reject H0
p-value
0.0051
Left Tailed Test
Right Tailed Test
CV
-2.68
CV
2.68
Decision DonotrejectH0
Decision Reject H0
p-value
0.0025
p-value
0.0025
Step 5:
since test stat (3.42) is less than CV (2.68), the null is rejected
We can conclude that the population mean fare has increased a
H1: μ
＞
267

T-Test One Sample test of hypothesis
288.30769 mean
22.45879 SD
Step 1
Given Data
Hypothesized (H0) pop mean mu
sample mean
sample size
sample SD
sig level alpha
degrees of freedom
Calculated Test-Statistic
T score
Two tailed test
CV
Decision
p-value
Left Tailed Test
CV
Decision
p-value
Step 5:
and there is sufficient evidence to support it.

Question # 48
pop mean, pop SD unknown
H0: μ = 27,000
H1: μ ≠ 27,000
60000
57000
79
12500
0.02
78
-2.13316666015574
2.38
-2.38
DonotrejectH0
0.036055240972422
Right Tailed Test
-2.0887
CV
2.0887
Reject H0
Decision DonotrejectH0
0.0180
p-value
0.0180
Since test stat (-2.13) is less than CV(2.38) the null is not rejected
We can conclude that the population mean is less than 60,000.

H0: μ = 0.63
H1: μ ≠ 0.63
Question # 50
Hypothesis Testing, One sample Proportion
Step 1:
H0: μ = 0.63
H1: μ ≠ 0.63
Given Data
Hypothesized pop. Proportion pi
60,000
sample proportion
57,000
sample size
79
12,500
sig level alpha
0.02
Calculated test statistic
z score
Err:502
Two Tailed Test
CV
2.33
Decision
Err:502
p-value
Err:502
Left-tailed Test
CV
-2.05
Decision
Err:502
p value
Err:502
Step 5:
Since test stat (2.51) is greater than CV(
We can conclude that the population p

n*pi>5
1
n*(1-pi)>5
0
0.99 adj. for ND
-2.33
Right Tailed test
CV
2.05
Decision
Err:502
p value
Err:502
V(1.96) the null is rejected
proportion is not less than 0.63.