2) The results of a 2012 Pew Foundation survey of - id220126160894, lealiastentz532542

Ответ:

Stetson02

24.01.2022

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:

Explanation:-

Step:- (1)

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

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

Step :-(ii)

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

H₀ : p₁ = p₂

Alternative hypothesis :H₁

H₁ : p₁ ≠ p₂

Level of significance : ∝=0.05

The tabulated value z=1.96

Step:-(iii)

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.

Conclusion:-

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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