A survey was given to a random sample of 1350 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. Of those surveyed, 26% of the people said they were in favor of the plan. Determine a 95% confidence interval for the percentage of people who favor the tax plan, rounding values to the nearest tenth.

The 95% confidence interval for the percentage of people who favor the tax plan is (23.7%, 28.3%).

Step-by-step explanation:

Confidence interval for the proportion:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

95% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

For the percentage:

Multiply the proportions by 100%.

0.237*100 = 23.7%.

0.283*100 = 28.3%.

The 95% confidence interval for the percentage of people who favor the tax plan is (23.7%, 28.3%).

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step-by-step explanation:

welll its a very hard question but ill

step-by-step explanation:

first you take the 6 dollars and add 10 to it then add the 50 to that then multiply by 8 with xy and your answer will be

hope this