100 points STATISTICS PLEASE HELP SOON Two opposing opinions were shown to a random sample of 2,150 buyers of a particular political news app in the United States. The opinions, shown in a random order to each buyer, were as follows:

Opinion A: The issue of equal pay is more important than increasing the minimum wage.
Opinion B: Increasing the minimum wage is more important than the issue of equal pay.

Buyers were to choose the opinion that most closely reflected their own. If they felt neutral on the topics, they were to choose a third option of "Neutral."

The outcomes were as follows:
50% chose Opinion A, 42% chose Opinion B, and 8% chose "Neutral."

Would a two-sample z-interval for a difference between proportions be an appropriate procedure to find if the difference in proportions between US buyers who would have chosen Opinion B and US buyers who would have chosen Opinion A is statistically significant? Explain why or why not.

A) 98% Confidence interval for the proportion of all US buyers of this particular app who would have chosen Opinion B

= (0.51, 0.57)

This means that the true proportion of all thay would chose opinion B can take on values between the range of (0.51, 0.57)

B) For the confidence interval obtained to be valid, the conditions stated must be satisfied and for the sampling distribution to be approximately normal, the number of buyers that chose Opinion B and the number of buyers that did not choose Opinion B must both be greater than 10.

Step-by-step explanation:

Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion of all US buyers of this particular app who would have chosen Opinion B = 0.54

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 98% confidence interval for sample size of 1500 is obtained from the z-tables.

Critical value = 2.33

Standard error = σₓ = √[p(1-p)/n]

p = sample proportion = 0.54

n = sample size = 1500

σₓ = √(0.54×0.46/1500) = 0.0128685664 = 0.01287

98% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.54 ± (2.33 × 0.01287)

CI = 0.54 ± 0.02998)

98% CI = (0.51, 0.57)

98% Confidence interval = (0.51, 0.57)

B) The number of buyers that chose Opinion B and the number of buyers that did not choose Opinion B are both greater than 10. Why must this inference condition be met?

For this confidence interval to be obtained using sample data, a couple of conditions are necessary to be satisfied. They include;

- The sample data must have been obtained using a random sampling technique.

- The sampling distribution must be normal or approximately normal.

- The variables of the sample data must be independent of each other.

On the second point, the condition for a binomial distribution to approximate a normal distribution is that

np ≥ 10

and n(1-p) ≥ 10

The quantity np is the actual sample mean which is the actual number of buyers that chose Opinion B while n(1-p) is the number of buyers that did not chose Opinion B.

For the confidence interval obtained to be valid, the conditions stated must be satisfied and for the sampling distribution to be approximately normal, the number of buyers that chose Opinion B and the number of buyers that did not choose Opinion B must both be greater than 10.

Hope this Helps!!!

y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead