Question

We draw a random sample of size 36 from a population with standard deviation 3.5. If the sample mean is 27, what is a 95% confidence interval for the population mean?

Answer 1

Answer:

The 95% confidence interval for the population mean is between 25.86 and 28.14.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{3.5}{\sqrt{36}} = 1.14$

The lower end of the interval is the sample mean subtracted by M. So it is 27 - 1.14 = 25.86

The upper end of the interval is the sample mean added to M. So it is 27 + 1.14 = 28.14

The 95% confidence interval for the population mean is between 25.86 and 28.14.

Answer 2

The 95% confidence interval for the population mean is between 25.86 and 28.14

mean (μ) = 27, standard deviation (σ) = 3.5, sample size (n) = 36

Confidence (C) = 95% = 0.95

α = 1 - C = 1 - 0.95 = 0.05

α/2 = 0.025

The z score of α/2 is the same as the z score of 0.475 (0.5 - 0.025) which is equal to 1.96.

The margin of error (E) is:

$E=z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } =1.96*\frac{3.5}{\sqrt{36} } =1.14$

The confidence interval = μ ± E = (27 ± 1.14) = (25.86, 28.14)

The 95% confidence interval for the population mean is between 25.86 and 28.14

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