A researcher estimates the 90% CI for a sample with a mean of M = 8.9 and a standard error (sigma_M) of 2.1. What is the confidence interval at this level of confidence?

Respuesta :

Answer:

The confidence interval at this level of confidence is between 5.4455 and 12.3545.

Step-by-step explanation:

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

[tex]\alpha = \frac{1-0.9}{2} = 0.05[/tex]

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

So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]

Now, find M as such

[tex]M = z*s[/tex]

In which s is the standard deviation of the sample, which is also called standard error. So

[tex]M = 1.645*2.1 = 3.4545[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is 8.9 - 3.4545 = 5.4455

The upper end of the interval is the sample mean added to M. So it is 8.9 + 3.4545 = 12.3545

The confidence interval at this level of confidence is between 5.4455 and 12.3545.

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