Mathematics, 03.02.2020 01:56, dstyle

What is the z-score with a confidence level of 95% when finding the margin of error for the mean of a normally distributed population from a sample? 0.99 1.65 1.96 2.58

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answered: hiji0206

For a confidence interval of 95%, z is 1.96

Step-by-step explanation:

A z-score, also called as standard score, indicates how many standard deviations away an element is from the mean. z-score can be calculated from the following formula,

$z=\dfrac{X-\mu}{\sigma}$

where,

X = Raw score,

μ = mean,

σ = standard deviation.

For a confidence interval of 95%, we take z as 1.96

answered: elizabethivy75

1.96

Further explanation

We will answer what is the value of the Z-score with a confidence level of 95% when finding the margin of error for the mean of a normally distributed population from a sample.

Z represents the value from the standard normal distribution for the selected confidence level.

See below, Z-scores for commonly used confidence intervals.

Desired Confidence Interval = 90%, with Z-score = 1.645Desired Confidence Interval = 95%, with Z-score = 1.96Desired Confidence Interval = 99%, with Z-score = 2.576

Thus, the Z-score with a confidence level of 95% is $\boxed{\boxed{ \ Z = 1.96 \ }}$

Notes:

A confidence level of 95% (or confidence intervals) mean that if we take 100 different samples and calculate 95% confidence intervals for each sample, then about 95 out of 100 confidence intervals (CI) will contain the true mean value (μ).

In practice, we choose a random sample and produce a confidence interval, which may or may not contain the true mean. The observed interval may be more or less than μ. As a result, 95% CI represents the range of possible precise and unknown parameters.

The formulas for confidence levels for the population mean depend on the sample size and are given below.

Confidence Intervals for μ

For n ≥ 30, $\boxed{ \ X = \pm Z \bigg( \frac{s}{\sqrt{n}} \bigg) \ }$ . We use the Z table for the standard normal distribution.For n < 30, $\boxed{ \ X = \pm t \bigg( \frac{s}{\sqrt{n}} \bigg) \ }$ . We use the t table with df = n - 1

With,

μ = the population meann = the sample sizeX = the sample means = the standard deviationdf = degree of freedom

In general, the relationship between the margin of error and the sample size is as follows:

The margin of error will be very small due to the large sample size. The margin of error will be greater due to the small sample size.Learn moreEvaluate a combination and a permutation Which graph represents the sequence The derivatives of the composite function

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answered: Guest

step-by-step explanation:

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