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kobusy [5.1K]
3 years ago
7

Hilary put 5 items on his birthday wish list. He received 3 of the items for his birthday. What percent of the

Mathematics
1 answer:
Misha Larkins [42]3 years ago
8 0

Answer:

60%

Step-by-step explanation:

100 ÷ 5 = 20

20 × 3 = 60

60 = 60%

I hope this helps

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

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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

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