What is the product of 41/6 and 2 1/5
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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
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
<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
where
q = 1-p
In given data n₁ = n₂ = n
on calculation , we get p = 0.2655
q =1-p = 1-0.2655
q = 0.7345
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