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 samples. Calculate the P-value if the observed statistic is Round your answer to four decimal places (e.g. 98.7654).

Answer :

Answer:

The answers to the questions are

(a) P = 0.0176

(b)  P = 0.5

(c)  P = 0.0571

Step-by-step explanation:

The P value is the calculated probability of obtaining extreme reults which are as extreme as observed res ult when the study null hypothesis H₀ is true

The p value is given by

P ([tex]\overline{\rm x}[/tex] = 185 when μ = 175)

z =  [tex]\frac{ \overline { \rm x} -\mu}{\sigma /\sqrt{n} }[/tex]

z([tex]\overline { \rm x}[/tex] = 185)  = [tex]\frac{ 185 -175}{20/\sqrt{10} }[/tex] = 1.5811

Therefore the P ([tex]\overline{\rm x}[/tex] = 185 when μ = 175) =  P(z = 1.5811)  

    0.94295

The probability density function is

[tex]f(x) = \frac{1}{\sigma\sqrt{2\pi } } e^{-\frac{1}{2}(\frac{x-\mu}{\sigma} )^2 }[/tex] Therefore the probability is

[tex]f(x) = \frac{1}{20\sqrt{2\pi } } e^{-\frac{1}{2}(\frac{185-175}{20} )^2 }[/tex] = 0.0176  

b) The type II error is given by

β = P(type II error) = P(Fail to reject the null hypothesis when it is false)

Therefore

β = P ([tex]\overline { \rm x}[/tex] < 185 when μ = 185)

P[tex](\frac{ \overline { \rm x} -\mu}{\sigma /\sqrt{n} } <\frac{185-185}{20/\sqrt{10} } )[/tex], which gives

P(z<0) and β = 0.5

(c) For true mean foam height of 195 mm we have

β = P ([tex]\overline { \rm x}[/tex] < 185 when μ = 195)

and P[tex](\frac{ \overline { \rm x} -\mu}{\sigma /\sqrt{n} } <\frac{185-195}{20/\sqrt{10} } )[/tex]

Which gives P(z < -1.58)

From z table the probability is       β = 0.0571

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