Question

People who experience online harassment either ignore it or respond via postings or involving authorities Is one method more effective than the other? A 2014 Pew Research Center report, based on a random sample of American adults, found that 412 of the 549 respondents who had chosen to ignore online harassment thought that it was an effective way to deal with the issue. Of the 368 respondents who had chosen to respond to online harassment, 305 thought that it had been effective A.

Required:
a. Is there good evidence that one approach is more likely to be effective than the other? State the null and alternative hypotheses, obtain the test statistic and P-value, and conclude in context.
b. How large is the difference? Obtain and interpret in context a 95% confidence interval for the difference between the proportions of American adults who find ignoring online harassment effective and those who find responding to it effective.
c. How effective is that best method? Obtain and interpret in context a 95% confidence interval for theproportion of American adults who find the best m ethod effective.

Answer 1

To determine the effectiveness of methods for handling online harassment, we use a two-proportion z-test to calculate the P-value and test the null hypothesis of no difference. We can calculate a 95% confidence interval for the difference in proportions and for the effectiveness of the best method.

Step-by-step explanation:

To assess whether one method is more effective than the other in handling online harassment, we look at the proportion of respondents who found each method effective. We calculate the proportion for ignoring harassment as 412/549, and for responding to it as 305/368. We can then set up the null hypothesis (H0) that there is no difference in effectiveness and the alternative hypothesis (H1) that there is a difference.

To test the hypothesis, we can use a two-proportion z-test. The test statistic is calculated using the difference in sample proportions and the standard error of the difference in proportions. The P-value corresponding to the test statistic will be computed from the standard normal distribution. If the P-value is smaller than the chosen significance level (commonly 0.05), we reject the null hypothesis.

For part b, we calculate a 95% confidence interval for the difference between the two proportions to understand how large the difference is. This interval will provide a range within which we can say the true difference in proportions lies with 95% confidence.

For part c, we calculate a 95% confidence interval for the proportion found for the best method. This will provide a range within which the true proportion of the effectiveness of the best method lies with 95% confidence.

Answer 2

Step-by-step explanation:

First method information:

Sample size n₁ = 549

x₁ ( people ignoring on-line harassment ) = 412

Proportion of people ignoring on-line harassment p₁

p₁ = 412/549 p₁ = 0,75 q₁ = 1 - p₁ q₁ = 0,25

Second method ( people chosen to respond)

Sample size n₂ = 368

number of people responding x₂ = 305

Proportion of people responding

p₂ = 305/ 368 p₂ = 0,83 then q₂ = 0,17

a) Looking for differences in the approachs

Hypothesis test:

Null Hypothesis H₀ p₁ = p₂

Alternative Hypothesis Hₐ p₁ ≠ p₂

Alternative Hypothesis indicates that the test is a two-tail test

Choosing a confidence interval CI = 95 % then a significance level is α = 5% α = 0,05 and from z-table we get z(c) = 1.64

To calculate z(s)

z(s) = ( p₁ - p₂ ) / EED

EED = √ p₁*q₁)/n₁ + (p₂*q₂)n₂

EED = √ (0,75*0,25)/549 + 0,83*0,17)/368

EED = √ 3,41 *10⁻⁴ + 3,83*10⁻⁴

EED = 10⁻² *√ 3,41 + 3,83

EED = 10⁻² * 2,69

EED = 0,0269

z(s) = ( p₁ - p₂ )/0,0269

z(s) = - 0,08 / 0,0269

z(s) = - 2,97

p-value for z(s) is pvalue = 0,0015

Comparing z(s) and z(c) modules

z(s) > z(c) therefor z(s) is in the rejection region we have to reject H₀ and claim that there is a difference between the proportions

b) How large is the difference