Which critical value is appropriate for a 99% confidence level where n = 17; Η is unknown and the
population appears to be normally distributed?

Respuesta :

Answer:

The critical value is ±2.921.

Step-by-step explanation:

The (1 - α) % confidence interval for population mean for unknown population standard deviation (σ) is:

[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\ \frac{s}{\sqrt{n}}[/tex]

The information provided is:

n = 17

Confidence level = 99%

Compute the value of α as follows:

(1 - α) % = 99%

     1 - α = 0.99

         α = 0.01

The degrees of freedom of the t-value is:

degrees of freedom = n - 1

                                  = 17 - 1

                                  = 16

Compute the critical value of t as follows:

[tex]\text{Critical value}=t_{\alpha/2, (n-1)}[/tex]

                     [tex]=t_{0.01/2, 16}\\=t_{0.005, 16}\\=\pm 2.921[/tex]

*Use a t-table for the critical value.

Thus, the critical value is ±2.921.

Ver imagen warylucknow

Using the t-distribution, it is found that the appropriate critical value is t = 2.9208.

The standard deviation for the population is unknown, hence, the t-distribution is used.

To find the critical value, two parameters are needed.

  • The confidence level.
  • The number of degrees of freedom(df), which is one less than the sample size.

For this problem, there is a confidence level of 99% and 17 - 1 = 16 df, hence, using a t-distribution calculator, the critical value is t = 2.9208.

A similar problem is given at https://brainly.com/question/15180581