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Makovka662 [10]
3 years ago
14

Aiden is trying to save money and has $1800 to set aside in some type of savings account. He checked his bank one day, and found

that the rate for a 12-month CD had an annual percentage yield of 4.25%. The interest rate on his savings account was 2.75% annual percentage yield. How much more simple interest wpuld Aiden earn if he invested in a CD for 12 months rather than leaving the $1800 in a regular savings account?
Mathematics
1 answer:
kipiarov [429]3 years ago
8 0
I = Prt

Interest on CD = 0.0425 x $1,800 = $76.5
Interest on savings = 0.0275 x $1,800 = $49.5

Difference of interest on CD and Interest on savings = $76.5 - $49.5 = $27

Therefre, he will earn $27 more if he invest in a CD rather than in a savings.
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Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

       = 3 - 1 = 2

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        = 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

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$\overline X_1 =50.5$           $\overline X_2 =50.07143$        $\overline X_3 =55.71429$

$s_1^2=19.96154$      $s_2^2=14.68681$         $s_3^2=36.57143$

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        = 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

     $=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

   $=\frac{127.1143}{2}$

   = 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

    $=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

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The F-test  statistics value is :

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  $=\frac{63.55714}{20.93304}$

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So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

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Between    127.1143                 2      63.55714          3.036212

Within        669.8571             32      20.93304

Total           796.9714            34

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