A survey was given to a random sample of 1450 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. Of those surveyed, 48% 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.
1 answer:
Answer:
(0.4543 ; 0.5057)
Step-by-step explanation:
This is a confidence interval for one sample proportion :
Phat , = 0.48
Sample size = 1450
1 - phat = 1 - 0.48 = 0.52
Zcritical = 95% ; Zcritical at 95% = 1.96
Using the relation :
Phat ± Zcritical *√[phat(1-phat) / n]
√[phat(1-phat) / n] = √(0.48*0.52) /1450) = 0.0131201
Lower boundary :
0.48 - (1.96 * 0.0131201) = 0.4543
Upper boundary :
0.48 + (1.96 * 0.0131201) = 0.5057
(0.4543 ; 0.5057)
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I hope this is good enough:
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