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g100num [7]
3 years ago
15

The following data values represent a population. What is the variance of the values? 7, 15, 11, 1, 11 A. 32 B. 18 C. 8 D. 22

Mathematics
1 answer:
Dominik [7]3 years ago
6 0
First, we find the mean....
(7 + 15 + 11 + 1 + 11) / 5 = 45 / 5 = 9

now we subtract the mean from every data point, and square it...
7 - 9 = -2.....-2^2 = 4
15 - 9 = 6....6^2 = 36
11 - 9 = 2....2^2 = 4
1 - 9 = -8...-8^2 = 64
11 - 9 = 2...2^2 = 4

now we find the mean of those numbers...that is ur varience...
(4 + 36 + 4 + 64 + 4) / 5 = 112/5 = 22.4....rounds to 22 <==
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Serggg [28]

Answer:

The calculated value z =  9.4451 > 1.96 at 5% level of significance.

Null hypothesis is rejected at 5% level of significance

yes there is  difference in the distribution of types of cell phones for the teachers in 2018 at a 5% level of significance

Step-by-step explanation:

<u>Explanation</u>:-

<u>Step:- (1)</u>

The results of a 2012 Pew Foundation survey of high school and middle school teachers is given in the pie chart.

A student asked a random sample of teachers in 2018 and found 165 had smart-phones, 80 had a cell phone

The first sample proportion

                                  p_{1} = \frac{80}{165} = 0.4848

A student asked a random sample of teachers in 2018 and found 165 had smart-phones,5 had no cell phone

The second sample proportion

                                p_{2} = \frac{5}{165} = 0.03030

<u>Step :-(ii)</u>

<u>Null hypothesis :H₀</u>: Assume that there is no difference in the distribution of types of cell phones for the teachers in 2018

H₀ : p₁ = p₂

<u>Alternative hypothesis :H₁</u>

H₁ : p₁ ≠ p₂

<u>Level of significance : ∝=0.05</u>

The tabulated value z=1.96

<u>Step:-(iii)</u>

The test statistic

                       Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }  } +\frac{1}{n_{2} } )} }

   where p = \frac{n_{1}p_{1} +n_{2} p_{2}  }{n_{1} + n_{2} }

              q = 1-p

 In given data n₁ = n₂ = n

              p = \frac{165 (0.4848)+165 (0.03030  }{165 + 165}

   on calculation , we get      p =  0.2655

                                                 q =1-p = 1-0.2655

                                                  q = 0.7345

                 

                    Z = \frac{0.4848 -0.030}{\sqrt{0.2655X0.7345(\frac{1}{165 }  } +\frac{1}{165} )} }

                    Z =   9.4451

The calculated value z =  9.4451 > 1.96 at 5% level of significance.

<u>Conclusion:</u>-

Null hypothesis is rejected at 5% level of significance

yes there is  difference in the distribution of types of cell phones for the teachers in 2018 at a 5% level of significance

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Step-by-step explanation:

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I hope this helps!

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