Answer:
d. Approximate the standard normal distribution with the Student's t distribution
(0.2199 ; 0.2327)
Step-by-step explanation:
Given that :
Sample size, n = 31
Sample mean, xbar = 0.2258
Sample standard deviation, s = 0.0188
Confidence interval (C. I) :
xbar ± margin of error
Margin of Error : Tcritical * s/sqrt(n)
Degree of freedom, df = n - 1 = 31 - 1 = 30
Tcritical value :
T0.05/2, 30 = 2.042
Margin of Error = 2.042 * 0.0188/sqrt(31)
Margin of Error = 0.0068949
C. I = 0.2258 ± 0.0068949
Lower boundary : (0.2258 - 0.006895) = 0.2189
Upper boundary : (0.2258 - 0.006895) = 0.2327
(0.2199 ; 0.2327)
<span>2.5(−4.4 − 3.5)
=</span>−4.4 − 3.5=-7.9
=2.5(-7.9)
=-19.75
Answer:
$3.55
Step-by-step explanation:
You need to add 29% of $2.75 to $2.75.
That means the price will be 129% of $2.75.
129% of $2.75 =
= 1.29 * $2.75
= $3.55
In my opinion I would pick the top for both answers
