EBK THE BASIC PRACTICE OF STATISTICS
EBK THE BASIC PRACTICE OF STATISTICS
8th Edition
ISBN: 8220106747841
Author: Moore
Publisher: YUZU
Question
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Chapter 18, Problem 18.49E

(a)

To determine

To find: The z statistic in terms of the sample mean x¯ .

To find: The value of z does the two-sided test reject H0 at the 5% level of significance.

(a)

Expert Solution
Check Mark

Answer to Problem 18.49E

The value of z statistic in terms of the sample mean x¯ is 0.894x¯8.944 .

The two-sided test will reject H0 if z>1.96 or z<1.96 .

Explanation of Solution

Given info:

The population of all measurements is normal with standard deviation σ=2.5 .

Calculation:

z statistic:

Formula for z statistic is z=x¯μ0(σn) .

Substitute 10 for μ0 , 2.5 for σ and 5 for n,

z=x¯10(2.55)=x¯1.118101.118=0.894x¯8.944

The value of z statistic is 0.894x¯8.944 .

Rejection rule for two-sided hypothesis:

If |z|>z* , then reject the null hypothesis (H0) .

From table C: t Distribution Critical Values, the value of z* corresponding with α=0.05 and two-sided hypothesis is 1.96. That is, z*=1.96 .

Thus, the two-sided test will reject H0 if z>1.96 or z<1.96 .

(b)

To determine

To find: The values of x¯ lead to rejection of H0 .

(b)

Expert Solution
Check Mark

Answer to Problem 18.49E

The values x¯>12.197 or x¯<7.812 will lead to rejection of H0 .

Explanation of Solution

Calculation:

Here, the two-sided test will reject H0 if z>1.96 or z<1.96 .

Consider,

z>1.96 .

Substitute 0.894x¯8.944 for z,

z>1.960.894x¯8.944>1.960.894x¯>10.904x¯>12.197

Consider,

z<1.96 .

Substitute 0.894x¯8.944 for z,

z<1.960.894x¯8.944<1.960.894x¯<6.984x¯<7.812

Thus, the values x¯>12.197 or x¯<7.812 will lead to rejection of H0 .

(c)

To determine

To find: The probability of observing x¯ that leads to rejection of H0 .

(c)

Expert Solution
Check Mark

Answer to Problem 18.49E

The probability of observing x¯ that leads to rejection of H0 is 0.4286.

Explanation of Solution

Given info:

The value of μ is 12.

Calculation:

Power:

Formula for power is,

Power=P(x¯>12.197)+P(x¯<7.812)=P(x¯μ(σn)>12.197μ(σn))+P(x¯μ(σn)<7.812μ(σn))=P(Z>12.19712(2.55))+P(Z<7.81212(2.55))=P(Z>0.1971.118)+P(Z<4.1881.118)

              =P(Z>0.18)+P(Z<3.75)=1P(Z0.18)+P(Z<3.75)

From Table A: Standard Normal Cumulative Proportions, the value of P(Z0.18) is 0.5714 and the value of P(Z<3.75) is 0.0000.

Therefore,

Power=1P(Z0.18)+P(Z<3.75)=10.5714+0.0000=0.4286

Thus, the probability of observing x¯ that leads to rejection of H0 is 0.4286.

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