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In spring of 2018 the Pew Research Council surveyed a panel of Russian citizens, asking them...

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%, and the format given in class. 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%?

math   Statistics-And-Probability   
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Answei Date: 30/01/2019 The point estimate of population proportion is, 359 506 0.7095 The point estimate of populationproportion is 0.7095 A 95% confidence interval for the proportion of Russian citizens who thought that their government did not try to intervene in the 2016 US election is, From the standard normal table, the z-critical value at 95% confidence is 1 .96.p(-p) 0.70951-0.7095) 506 = 0.7095 ± 1.96 × 0.6531 to 0.7659 A 95% confidence interval for the proportion of Russian citizens who thought that their government did not try to intervene in the 2016 US election is 0.6531 to 0.7659 Interpretation of confidence interval There are 95% confident that the proportion of Russian citizens who thought that their government did not try to intervene in the 2016 US election is between 0.6531 and 0.7659 To test the hypothesis is that the percentage of Russian citizens who thought that their government did not try to intervene in the 2016 US election is not 50% at 5% level of significance.The nulland alternative hypothesis is Ho : p = 0.50 H, : p # 0.50 The value of test statistics is, p-p P(l-p) 0.7095 -0.50 0.50(1 0.50) 506 = 9 .42 The value of test statistics is 9.42 The z-critical value is,From the standard normal table, the z-critical value for two-tailed at 95% confidence is 1.96 The z-critical value is 1.96 Decision The z-test statistics which is 9.42 corresponds to sample statistics is falls in the critical region which is 1.96, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the percentage of Russian citizens who thought that their government did not try to intervene in the 2016 US election is not 50%. The result is statistically significant.

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