Show that for a sample of n = 39, the smallest and largest Z values are – 1.96 and + 1.96, and the middle (i.e., 20th) Z value is XXXXXXXXXXThe data in the file spending represent the per-capita...

Solutions

Solution 13.1: ̅

a) When X = 0, the estimated expected value of Y is 2.The intercept means that
the Y is increasing independent of number of units of X.
b) There exists a positive relationship between the number of guest
establishments and the number of guests nights stayed ('000) used to predict
the tourist behavior. For each increase in the value X by 1 unit, you can
expect an increase by an estimated 5 units in the value of Y.
c) Predicting the mean value of Y for X = 3
̅
̅

Thus, the mean value of Y is 17 units when X is 3 units.

Solution 13.2:
a) Yes
b) No
c) No
d) Yes

Solution 13.4:
a)


b) For these data, ̅
bo=145 and b1=7.4.
For each increase in shelf space of an additional foot, there is an expected
increase in weekly sales of an estimated 0.074 hundreds of dollars, or $7.40.
c) Predicting the mean weekly sales of pet food for stores with 8 feet of shelf
space for pet food.
̅
̅ 0
Thus, the mean value of Y is $204.20 for stores with 8 feet of shelf space.

Solution12.1:
a) For degrees of freedom = 1 and α = 0.01, χ2 = 6.6349
b) For degrees of freedom = 1 and α = 0.025, χ2 = 5.0239.
c) For degrees of freedom = 1 and α = 0.05, χ2 = 3.8415.
Solution12.3:
(a) – (b)
Observed Freq Expected freq
20 20
Chi-sq contribution = 0
Observed Freq Expected freq
30 30
Chi-sq contribution = 0
Total Obs, Row 1
50
Observed Freq Expected freq
30 30
Chi-sq contribution = 0
Observed Freq Expected freq
45 45
Chi-sq contribution = 0
Total Obs, Row 2
75
Total Obs, Col 1
50
Total Obs, Col 1
75
Total Obs, Col 1
125
c) ∑
( )



Since it is not significant at the 5% level of significance.
















Solution12.5:
a)
Observed Frequencies
Gender
Enjoy Shopping Male Female Total
Yes 136 224 360
No 104 36 140
Total 240 260 500
Expected Frequencies
Gender
Enjoy Shopping Male Female Total
Yes 172.8 187.2 360
No 67.2 72.8 140
Total 240 260 500

Gender
Level of Significance 0.01
Number of Rows 2
Number of Columns 2
Degrees of Freedom 1
Results
Critical Value 6.634891
Chi- Square test
statistic
53.8258
p- value 2.19E - 13
Reject the null hypothesis

H0: π1 = π2 Ha: π1 ≠ π2 where population: 1 = males and 2 = females
Decision rule: df = 1, if χ2 = 6.635, reject H0
Test statistic: χ2 = 53.8258
Decision: Since χ2calc = 53.8258 is greater than the upper critical bound of 6.6349,
reject H0. There is enough evidence to conclude that there is significant difference
between the proportions of males and females who enjoy shopping for clothing at
the 0.01 level of significance.
b) p-value = virtually zero. The probability of obtaining a test statistic of 53.8258 or
larger when the null hypothesis is true is virtually zero.
c) (a) H0: π1 = π2 Ha: π1 ≠ π2 where population: 1 = males and 2 = females
Decision: Since χ2calc = 0.0106 is less than the upper critical bound of 6.635, do
not reject H0. There is not enough evidence to conclude that the proportion of
males and females who enjoy shopping for clothing are different.
(b) p-value = 0.9180. The probability of obtaining a test statistic of 0.0106 or
larger when the null hypothesis is true is 0.9180.
d) The results of (a) – (c) are exactly the same as those of Problem 10.31. The χ2calc
in (a) and the (Zcalc)
2 in Problem 10.31 (a) satisfy the relationship that χ2calc =
53.8258 = (Zcalc)
2 = (-7.34)2 and the p-value obtained in (b) is exactly the same as
the p-value in Problem 10.31 (b). Similarly, the χ2calc in (c) and the Zcalc in Problem
10.31 (d) satisfy the relationship that χ2calc = 0.0106 = (Zcalc)
2 = (0.103)2 and the p-
value obtained in (c) is exactly the same as the p-value in Problem 10.31 (d).
Solution12.11:
a) df = (r – 1)(c – 1) = (2 – 1)(5 – 1) = 4
b) χ2= 9.488
c) χ2= 13.277

Solution12.12:
a) The expected frequencies in the first row are 20, 30, and 40.
The expected frequencies in the second row are 30, 45, and 60.
b) χ2= 12.500. The critical value with 2 degrees of freedom and α = 0.05 is 5.991.
The result is deemed significant.
c) Pairs of proportions that differ at the 0.05 level are marked with * below:

Pairwise Comparisons Critical Range |pj – pj’|
A to B 0.19582 0.2*
A to C 0.1848 0.3*
B to C 0.1848 0.1
There are two (2) pairs of proportions that differ significantly.






Solution12.14:
a)
Number of Categories: 5
Degrees of freedom: 4
Expected Freq: 29.2

Test Statistic, χ2 = 52.5616
Critical χ2 = 9.48772
P-Value: 0.0000

Note that the test statistic X^2 = Sum (O - E)^2 / O is much bigger than the Critical
Value so we are far in the rejection region.

H0: π1 = π2 = π3 = π4 = π5 Ha: Not all π are equal
Where population 1 = Germany, 2 = France, 3 = UK, 4 = Greece, 5 =US

Decision: Reject the Null Hypothesis and conclude that there is a difference in the
proportion of people who eat out at least once a week in the various countries.

b) P-value is practically zero, which is for sure less than any alpha (= level of
significance)

You can establish USA and Germany are both different from the rest. (Way
different from 29.2%)

c) Excel output of the Marascuilo procedure:

Sample Sample
Group Proportion Size
1 0.10 1000
2 0.12 1000
3 0.28 1000
4 0.39 1000
5 0.57 1000
Other Data
Level of
Significance
0.05
df 4
Sqrt(Chi-
square)
3.0802

Absolute
Difference
Std. Error Of
Difference
Critical Range Results
Comparison
Grp1 to Grp2 0.02 0.01399 0.04308 Means are not different
Grp1 to Grp3 0.18 0.01708 0.05260 Means are different
Grp1 to Grp4 0.29 0.01811 0.05578 Means are different
Grp1 to Grp5 0.47 0.01831 0.05639 Means are different
Grp2 to Grp3 0.16 0.01753 0.05399 Means are different
Grp2 to Grp4 0.27 0.01853 0.05709 Means are different
Grp2 to Grp5 0.45 0.01873 0.05768 Means are different
Grp3 to Grp4 0.11 0.02096 0.06457 Means are different
Grp3 to Grp5 0.29 0.02114 0.06510 Means are different
Grp4 to Grp5 0.18 0.02198 0.06769 Means are different
At 5% level of significance, there is no significant difference between the
proportions of Germany and France while there is significant difference between all
the remaining pair of countries.

Solution11.1:
(a) Degrees of freedom under among group...