A consumer products company is formulating a new shampoo and is interested in foam height (in millimeters). Foam height is approximately normally distributed and has a standard deviation of 20 millimeters. The company wishes to test millimeters versus millimeters, using the results of n samples. Find the boundary of the critical region if the type I error probability is and

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Complete question:

A consumer products company is formulating a new shampoo and is interested in foam height (in millimeters). Foam height is approximately normally distributed and has a standard deviation of 20 millimeters. The company wishes to test H0: u=175 millimeters versus Ha:u>175 millimeters, using the results of n samples. Find the boundary of the critical region if the type I error probability is [tex] \alpha = 0.01 [/tex] and n = 16

Answer:

186.63

Step-by-step explanation:

Given:

[tex] \alpha = 0.01 [/tex]

Using the standard normal deviate table:

NORMSINV(0.01) = 2.326

Thus, the Z score = 2.326

To find the critical value if the mean, use the formula:

[tex]\frac{X' - u_0}{\sigma/\sqrt{n}} = Z[/tex]

Since we are to find X', Make X' subject of the formula:

[tex] X' = u_0 + (Z * \frac{\sigma}{\sqrt{n}}) [/tex]

[tex] X' = 175 + (2.326 * \frac{20}{\sqrt{16}}) [/tex]

[tex] X' = 175 + (2.326 * 5) [/tex]

[tex] X' = 175 + 11.63 [/tex]

[tex] X' =186.63 [/tex]

The boundary of the critical region is 186.63