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Alisiya [41]
2 years ago
8

n studying the sampling distribution of the mean, you were asked to list all the different possible samples from a small populat

ion and then find the mean of each of them. Consider the following: Personal phone calls received in the last three days by a new employee were 2, 4, and 7. Assume that samples of size 2 are randomly selected with replacement from this population of three values. What different samples could be chosen? What would be their sample means?
Mathematics
1 answer:
Mashutka [201]2 years ago
8 0

The Total outcome with replacement will be given by ,3² = 9, the mean of the samples will be  2 , 4.5 , 3 , 7 , 4

<h3>What is Mean ?</h3>

Mean predicts the central tendency of a data set , It is determined by the ratio of sum of all the numbers to the total numbers in the data set.

On the basis of given data

The number given is 2,4,7

Size of the sample = 2 with replacement

The Total outcome with replacement will be given by

3² = 9

Possible outcome can be

2,2   2,4   2,7   4,4  4,2  4,7  7,7  7,2  7,4

The sample mean can be calculated as

(a+b)/2

(2+2)/2 = 2

similarly the mean of the samples will be  2 , 4.5 , 3 , 7 , 4

To know more about Mean

brainly.com/question/521501

#SPJ1

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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

d.f.D = N - k

        = 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

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

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

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

         $=\frac{51+51+...+49+49}{35}$

        = 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

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

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

   $=\frac{669.8571}{32}$

  = 20.93304

The F-test  statistics value is :

$F=\frac{s_B^2}{s_W^2}$

  $=\frac{63.55714}{20.93304}$

  = 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

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

The ANOVA table is :

Source       Sum of squares    d.f    Mean square    F

Between    127.1143                 2      63.55714          3.036212

Within        669.8571             32      20.93304

Total           796.9714            34

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