# interval deposits

```13.6.
a.
x,y Scatterplot
2.5
2
y
1.5
1
0.5
0
0
1
2
3
4
5
x
There appears to be a positive linear relationship.
b. Using Excel’s CORREL function, the sample correlation coefficient is 0.9239.
The correlation coefficient indicates the strength of the linear relationship
between two variables. In this case the relationship is a positive one.
c. H0: ρ &lt; 0
HA: ρ &gt; 0
d.f. = 10 - 2 = 8
Decision Rule:
If t &gt; 2.8965, Reject H0, otherwise do not reject H0
t = 0.9239/ (1  0.92392) /(10  2) = 6.8295
Since 6.8295 &gt; 2.8965 reject H0 and conclude that the correlation coefficient is
greater than 0.
13.7
a.
There appears to be a fairly strong positive relationship.
b. Using Excel’s CORREL function, the sample correlation coefficient is 0.847065.
The correlation coefficient indicates the strength of the linear relationship
between two variables. In this case the relationship is a positive one.
c. H0: ρ = 0
HA: ρ ≠ 0
d.f. = 7-2 = 5
Decision Rule:
If t &gt; 2.5706 or t &lt; -2.5706, Reject H0, otherwise do not reject H0
t = 0.847065/ (1  0.8470652) /( 7  2) = 3.5638
Since 3.5638 &gt; 2.5706 reject H0 and conclude that the correlation coefficient is
not 0.
d. Because we rejected the null hypothesis we could have committed a Type I Error.
13.10.
a. The correlation coefficient in this problem indicates there could be a weak linear
relationship between account balance and the number of deposits and
withdrawals. This indicates the larger the number of deposits and withdrawals the
smaller the bank account balance.
b. H0: ρ = 0
HA: ρ ≠ 0
d.f. = 50 - 2 = 48
Students can use Excel’s TINV function to determine the critical t-value for alpha
= .10 and df = 48 = + 1.6772
Decision Rule:
If t &gt; 1.6772 or t &lt; -1.6772, Reject H0, otherwise do not reject H0
2
t = -0.23/ (1  (0.23) ) /(50  2) = -1.6374
Since t is not in the rejection region, do not reject H0 and conclude that there is
no correlation between account balance and the number of deposits and
withdrawals.
c. Since the null hypothesis was not rejected, we could have committed a Type II
error.
13.16.
a.
x,y Scatterplot
16
14
12
y
10
8
6
4
2
0
0
50
100
150
x
200
250
There is a strong negative relationship.
b. Using Excel’s CORREL function, the correlation coefficient is -.82901.
H0: ρ = 0
HA: ρ ≠ 0
d.f. = 9 -2 = 7
Decision Rule:
If t &gt; 2.3646 or t &lt; -2.3646, Reject H0, otherwise do not reject H0
t = -0.82901/ (1  ( 0.829012 )) /( 9  2) = -3.922
Since –3.922 &lt; -2.3646 reject H0 and conclude that there is a correlation
between x and y.
c.
The regression equation is ŷ = 19.75 - .08346(x). The slope coefficient is .08346. This means that as the x variable increases one unit that the y variable
decreases by .08346 units. The y intercept is 19.75. This means that when the x
variable is equal to 0, the y variable is equal to 19.75.
d.
H0: 1 = 0
HA: 1 ≠ 0
d.f. = 9 -2 = 7
Decision Rule:
If t &gt; 2.9979 or t &lt; -2.9979, Reject H0, otherwise do not reject H0
t = -0.08346/ .02128 = -3.922
Since –3.922 &lt; -2.9979 reject H0 and conclude that there is a a linear relationship
between x and y.
e. Since we rejected the null hypothesis we might have made a Type I error.
13.21.
r = 0.9963
H0: ρ = 0
HA: ρ ≠ 0
d.f. = 11-2 = 9
α = 0.05 (Note, alpha was not specified so students may choose a different level
which will affect the critical value.)
Decision Rule:
If t &gt; 2.2622 or t &lt; -2.2622, Reject H0, otherwise do not reject H0
t = 0.9963/ (1  0.99632) /(11  2) = 34.7774
Since 34.7774 &gt; 2.2622 reject H0 and conclude there is a correlation between the
computer inventory record and the actual level of inventory
13.22.
a. The slope coefficient, -10.12, is the average decrease in total dollar volume in
business this year for each 1 mile increase in the distance the customer is from
b. H0: β1 &gt; 0
HA: β1 &lt; 0
Decision Rule
If z &lt; -2.33 reject H0, otherwise do not reject H0
z = (-10.12-0)/3.12 = -3.2436,
Since –3.2436 &lt; -2.33 reject H0 and conclude the further a business is from the
corporate headquarters the smaller its sales are.
13.29.
a.
b. H0: 1 = 0
HA: 1 ≠ 0
The regression equation is 6.258 + 0.925(x)
Decision Rule:
If  &lt; .1; reject H0, otherwise do not reject.
Since p-value = 0.0000 &lt;  &lt; .1 do not reject the hypothesis and conclude that the
number of units is not linearly related to the time required.
c. The 90% confidence interval is0.92489  1.8331(0.0722) = 0.7925 ----- 1.0573; it
does contain the hypothesized value (= 1). Therefore, you could conclude the
13.34.
a. 95% confidence interval for the true regression slope
0.124 + 1.96(2.88); -5.5208 ----- 5.7688
A 95% confidence interval means that if many random samples were taken from
this population, and regression models determined, 95% of the calculated slope
coefficients would have values in the above interval. Since 0 is in the interval, it
also means the model is not significant.
b. No, because the confidence interval contains the value of 0 which means the true
regression slope could be 0 which would mean that age and speed are not linearly
related.
13.35.
a. ŷ = -4.7933 + 1.0488x
b. ŷ = -4.7933 + 1.0488(100) = 100.09
c. 100.09 + 1.8331(9.333) 1 
1 (100  115.0909) 2

; 82.2671 - - - - - 12
116936.3
117.9129
d. 100.09 + 1.8331(9.333)
1 (100  115.0909) 2

; 95.0939 - - - - - - 105.0861
12
116936.3
```