The Basic Practice of Statistics
The Basic Practice of Statistics
7th Edition
ISBN: 9781464142536
Author: David S. Moore, William I. Notz, Michael A. Fligner
Publisher: W. H. Freeman
Question
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Chapter 18, Problem 18.49E

a.

To determine

The z test statistics in terms of the sample mean x¯ .

The value of z for which two sided test reject H0 at 5% level of significance.

a.

Expert Solution
Check Mark

Answer to Problem 18.49E

The null hypothesis will be rejected for Z1.96_or Z1.96_ at 5% level of significance.

Explanation of Solution

Given info:

Random sample of size 5 is as follows:

6.477.5110.1013.639.91

The Null and Alternative hypothesis is given as:

The null hypothesis:

H0:μ=8

The alternative hypothesis:

Ha:μ8

Simple random sample of size n is 5 and standard deviation is σ=2 .

Calculation:

The formula for sample mean is,

X¯=iXin

Where,

  • X is the sum of all sample values.
  • n is the sample size.
  • X¯ is the sample mean.

The formula for the z statistic is,

z=X¯μσn

Where,

  • X¯ is the sample mean.
  • μ is the population mean.
  • σ is the population standard deviation.
  • n is the sample size.

Add all the sample observations to find iXi .

iXi=6.47+7.51+10.10+13.63+9.91=47.62

Substitute 47.62 for iXi and 5 for n in equation for sample mean to find X¯ .

X¯=47.625=9.524

Substitute 8 for μ , 9.524 for X¯ , 2 for σ and 5 for n in equation for z statistic to find z statistic.

z=9.524825=1.5240.894=1.740

The value of z statistic is 1.740.

At a 5% level of significance, the critical value of z for two tailed test is 1.96.

From the table of “The standard Normal Cumulative Proportions”, the critical value for the left tail test corresponding to an area of 0.25 is 1.96 and for the right tail test corresponding to an area of 0.975 (10.975=0.25) is 1.96.

Decision rule:

If absolute value of test statistic is less than critical value then there is not enough evidence to reject the null hypothesis.

The test statistic value 1.740 is less than that of the critical value 1.96, hence there is enough evidence to reject the null hypothesis.

Thus, null hypothesis will be rejected for Z1.96or Z1.96 .

b.

To determine

The value of sample mean lead to reject the null hypothesis.

b.

Expert Solution
Check Mark

Answer to Problem 18.49E

H0 is rejected when X¯6.247_andX¯9.753_ .

Explanation of Solution

Calculation:

The formula for the z statistic is given by;

z=X¯μσn (2)

Where,

  • X¯ is the sample mean.
  • μ is the population mean.
  • σ is the population standard deviation.
  • n is the sample size.

Substitute 8 for μ , 9.524, for σ and 5 for n in equation.

z=X¯825=52×(X¯8)=1.118×(X¯8)

Reject the null hypothesis H0 when,

=1.118(X¯8)1.96=(X¯8)1.961.118=X¯1.753+8=X¯9.753

Reject the null hypothesis H0 when,

=1.118(X¯8)1.96=(X¯8)1.961.118=X¯1.753+8=X¯6.247

Thus, at 5% level of significance level H0 is rejected when X¯6.247andX¯9.753 .

c.

To determine

The power of the test.

c.

Expert Solution
Check Mark

Answer to Problem 18.49E

The power of the test is about 39%.

Explanation of Solution

Calculation:

Power:

Power of a test of hypothesis is defined as the probability of rejecting null hypothesis when the alternative is true.

The sample mean X¯ follows normal distribution with population mean μ=10  and standard deviation σn=25 is defined as:

X¯N(10,25)

Therefore,

P(X¯6.247)+(X¯9.753)=P(6.247X¯9.753)=P(X¯μσnzX¯μσn)=P(6.2471025z9.7531025)=P(3.7530.8944z0.2470.8944)

=P(4.20z0.28)=P(z0.28)P(z4.20)=P(z>0.28)P(z>4.20)(Bysymmetric)=0.5P(z<0.28)(0.5P(z<4.20))

=P(z<4.20)P(z<0.28)=0.50.1103(UsingStandardNormalTableA)=0.3897

0.39

Hence, the power of the test is about 39%.

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