Question

In spring of 2018 the Pew Research Council surveyed a panel of Russian citizens, asking them whether they believed their government tried to influence the 2016 US Presidential Election. This survey used a complicated design - the open source statistics team produced a simplified version of this dataset which can be treated as a random, representative sample of independent observations of Russian citizens.

Tabulated responses to the question " Did your government tried to influence the 2016 election in the United States?" are given below

Responses	Frequency
"Did not try"	359
"Did try"	76
"Don't know"	71
Total number of respondents	506

i) Produce a point estimate, and an appropriate confidence interval estimate, for the proportion of Russian citizens who thought that their government did not try to intervene in the 2016 US election. Use a confidence level of 95%. Make sure you interpret your estimate.

ii) Do you think there is evidence that the percentage of Russian citizens who thought that their government did not try to intervene in the 2016 US election is not 50%?

Answer 1

a ) Produce a point estimate, and an appropriate confidence interval estimate, for the proportion of Russian citizens who thought that their government did not try to intervene in the 2016 US election. Use a confidence level of 95%. Make sure you interpret your estimate.

Point estimate for the proportion is sample proportion =p^ = # Did not try / total number of respondents

                                                                                             = 359/506 =0.7095.

Confidence interval for proportion formula :

$\hat{p}+/- E$    where $E = Z _{\alpha /2} * \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

So here Za/2 is the critical value at a/2 area . Here confidence level = 0.95

alpha = a= 1 -0.95 = 0.05 , a/2 = 0.05/2 = 0.025

Then in excel use command =NORMSINV(0.025) then hit enter you will get z-score value -1.96 ( round to two decimal ).

We are taking positive value of Za/2 = 1.96 .

n = total number of respondents = 506 , p^ = 0.7095 ,.Lets find value of E.

$E = Z _{\alpha /2} * \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} =1.96 * \sqrt{\frac{0.7095*(1-0.7095)}{506}} = 1.96 * \sqrt{\frac{0.2061}{506}}$

    $E = 1.96 * \sqrt{0.0004073}=1.96*0.020181=0.0396$

Lower limit : p^ -E = 0.7095- 0.0396 = 0.6699=67%

upper limit : p^ +E = 0.7095 +0.0396 = 0.7491=75%.

So here we conclude that at 95% confidence level for the proportion of Russian citizens who thought that their government did not try to intervene in the 2016 US election lies between 67% to 75%.

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Part 2)

Do you think there is evidence that the percentage of Russian citizens who thought that their government did not try to intervene in the 2016 US election is not 50%?

Claim : the percentage of Russian citizens who thought that their government did not try to intervene in the 2016 US election is not 50% : P $\neq$ 50% =0.50

Hypothesis : $H0:P=0.5$    vs    $H1:P \neq 0.5$.

Test statistics :

              $Z = \frac{\hat{p}-P}{\sqrt{\frac{P*(1-P)}{n}}}$ $= \frac{0.2095}{\sqrt{\frac{0.25}{506}}}=\frac{0.2095}{\sqrt{0.00049}}=\frac{0.2095}{0.02214}=9.4625$

So here Z test statistics = 9.46.

P- value : Here test is two tailed test so p- value =2*[ 1 - P(Z < 9.46 ) ]

              Use excel command , =2*(1-NORMSDIST(9.46)) then hit enter you will get answer for p- value as 4.39E-21

            this can be written in simple form as, 0.00000000000000000000439 ==>>0.000

So here p- value   < 0.05 ; reject H0.

Here we conclude that , at 5% level of significance , the percentage of Russian citizens who thought that their government did not try to intervene in the 2016 US election is not 50% .