sabrinaaz
18.07.2020 •
Mathematics
Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 315 with 37% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
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Ответ:
The 99% confidence interval is = 0.37 +/- 0.070
= (0.300, 0.440)
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
p+/-z√(p(1-p)/n)
Given that;
Proportion p = 37% = 0.37
Number of samples n = 315
Confidence interval = 99%
z value(at 99% confidence) = 2.58
Substituting the values we have;
0.37 +/- 2.58√(0.37(1-0.37)/315)
0.37 +/- 2.58√(0.00074)
0.37 +/- 2.58(0.027202941017)
0.37 +/- 0.070183587825
0.37 +/- 0.070
= (0.300, 0.440)
The 99% confidence interval is = 0.37 +/- 0.070
= (0.300, 0.440)
Ответ:
47.5
Step-by-step explanation:
3x - 10 + x = 180
4x = 190
x = 47.5