An important property of plastic clays is the percent of shrinkage on drying.A c

Question

An important property of plastic clays is the percent of shrinkage on drying.A certain type of plastic clay has a shrinkage that has a standard deviation of 1.2 percent.A sample of 38 results in an average shrinkage of 18.4 percent.

Test to determine whether the true average shrinkage is greater than 18 percent.Use alpha = .05.

What assumptions are required for this hypothesis test?

What is the power of the test, if the true mean is actually 18.5?

Please use Minitab for calculating the power if possible

Explanation / Answer

Here we have to test the hypothesis that,

H0 : mu = 18 Vs H1 : mu > 18

where mu is population average shrinkage.

Also population standard deviation (sigma) = 1.2

n = 38

Average shrinkage (Xbar) = 18.4

Alpha = 5% = 0.05

Here we use z-test for one sample.

Because population standard deviation is known.

This we can done using TI-83 calculator.

steps :

STAT --> TESTS --> 1:Z-Test --> ENTER --> High light on Stats --> ENTER --> Input mu0, sigma, Xbar and n --> select alternative ">mu0" --> ENTER --> Calculate --> ENTER

Test statistic Z = 2.0548

P-value = 0.0199

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion : There is sufficient evidence to say that the true average shrinkage is greater than 18 percent.

Power = 1 - type II error

Type II error = P(Accept H0 / H1 : mu = 18.5)

First we compute accptance region.

Critical value (0.05) = 1.645

Rejection region = RR = {Z / Z > 1.645 }

Accptance region = AR = {Z / Z < 1.645}

AR = { (Xbar - 18) / 1.2/sqrt(38) < 1.645 }

= { (Xbar - 18) / 0.1947 < 1.645 }

= { (Xbar - 18) > 0.3202 }

= { Xbar > 18.32}

Type II error = P(Xbar > 18.32 / H1 : mu = 18.5 }

= P { (Xbar - mu) / sigma / sqrt(n) > (18.32 - 18.5) / 0.1947 }

= P( Z > -0.92 )

This probability we can find bu using EXCEL.

syntax is,

= 1 - NORMSDIST(z)

where z is test statistic value.

Type II error = 0.8221

Power = 1 - 0.8221 = 0.1779

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