Answer :
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:
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
[tex]p_{1} = \frac{80}{165} = 0.4848[/tex]
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
[tex]p_{2} = \frac{5}{165} = 0.03030[/tex]
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
[tex]Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} } } +\frac{1}{n_{2} } )} }[/tex]
where [tex]p = \frac{n_{1}p_{1} +n_{2} p_{2} }{n_{1} + n_{2} }[/tex]
q = 1-p
In given data n₁ = n₂ = n
[tex]p = \frac{165 (0.4848)+165 (0.03030 }{165 + 165}[/tex]
on calculation , we get p = 0.2655
q =1-p = 1-0.2655
q = 0.7345
[tex]Z = \frac{0.4848 -0.030}{\sqrt{0.2655X0.7345(\frac{1}{165 } } +\frac{1}{165} )} }[/tex]
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