The sample space of a population being researched contains 58 values and a mean of 318.6. The population has a known standard deviation of 29.2. Identify the margin of error for a 95% confidence interval estimate of the mean of the population.

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.

x+/-zr/√n

x+/-M.E

Given that;

M.E = margin of error

Mean x = 318.6

Standard deviation r = 29.2

Number of samples n = 58

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

318.6+/-1.96(29.2/√58)

318.6+/-1.96(3.834147839503)

318.6+/-7.514929765427

318.6+/-7.51

Margin of error M.E = 7.51

The 95% Confidence interval is = 318.6+/-7.51

= (311.09, 326.11)

Cetacea Cetacea · Answer 2 · 2022-01-14T13:13:50+01:00

Cetacea Cetacea

14-01-2022

The margin of error is 7.5149.

Given that;

Mean = μ = 318.6

Standard deviation = σ = 29.2

Number of samples = n = 58

Confidence interval = 95%

z(at 95% confidence) = 1.96

We have to find M.E = margin of error = ?

[tex]z\cdot\frac{\sigma}{\sqrt{n}}\\=1.96\cdot\frac{29.2}{\sqrt{58}}\\=7.5149[/tex]

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