Please help I will mark you brainlist
1 answer:
You might be interested in
Answer:
The sample proportion is 0.88.
The margin of error is of 0.04.
The interval likely to contain the true population proportion is (0.84,0.92).
Step-by-step explanation:
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 z-score that has a p-value of .
Margin of error:
The margin of error is of:
Sample of 230 teenagers, 202 would like to have smaller class sizes at their school.
This means that , and that the sample proportion is:
The sample proportion is 0.88.
Standard 95% confidence level
So , z is the value of Z that has a p-value of
, so
.
Margin of error:
The margin of error is of 0.04.
Confidence interval:
Sample proportion plus/minus margin of error. So
0.88 - 0.04 = 0.84
0.88 + 0.04 = 0.92
The interval likely to contain the true population proportion is (0.84,0.92).