You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 7.
While it is an uncommon confidence level, find the critical value that corresponds to a confidence level of 95.9%.
(Report answer accurate to three decimal places with appropriate rounding.)
ta/2 = ±

Answer :

The critical value that corresponds to a confidence interval of 95.9% is 1.99.

The distribution of the sample mean will be roughly normally distributed, according to the Central Limit Theorem, if we have an unknown population with a mean and standard deviation and adequately small random samples (n < 30) are chosen from the population with replacement.

Then, the mean of the sample means is given by,

μ(x) = μ

And the standard deviation of the sample means is given by,

σ = σ / [tex]\sqrt{n}[/tex]

In this case the sample selected is of size, n = 7.

As the sample size n = 7 < 30, the sampling distribution of sample mean will be approximately normal.

So, a z-interval will be used to estimate the population mean.

The confidence level is, 95.9%.

The value of α is:

∝ = 1 - confidence level

∝ = 1 - 0.959

∝ = 0.041

The critical value is:

[tex]z_{\frac{∝ }{2} }[/tex] =  [tex]z_{\frac{0.041}{2} }[/tex]

[tex]z_{\frac{∝ }{2} }[/tex] = [tex]z_{0.0205}[/tex]

[tex]z_{\frac{∝ }{2} }[/tex] = -1.99

[tex]z_{1-∝/2}[/tex] = 1.99

Use a z-table.

Therefore,

The critical value that corresponds to a confidence interval of 95.9% is 1.99.

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