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Question: Find the following probabilities for the standard normal random variable z:

a.P(z>1.46)b.P(z<-1.56)c.P(.67≤z<2.41)d.P(-1.96≤z-.33)e.P(Z≥0)f.P(-2.33<z<1.50)

Short Answer

Expert verified

Answer

A random variable is a mathematical expression of a statistical study's result.

Step by step solution

01

(a) The data is given below

The calculation is given below:

P(z>1.46)=1-P(z≤1.46)=1-0.9279=0.0721

02

(b) The data is given below

The calculation is given below:

P(z<-1.56)=0.0594

03

(c) The data is given below

The calculation is given below:


P(.67≤z<2.41)=P(.67≤z<2.41)-P(z<.0.67)=0.99920-0.7486=0.2434

04

(d) The data is given below

The calculation is given below:

P(-1.96≤z-.33)=P(-1.96≤Z<0)-P(-0.33≤Z<0)=0.4750-0.1293=0.3457

05

(e) The data is given below

The calculation is given below:

P(Z≥0)=1-P(z<0)=1-0.5=.5

06

(f) The data is given below

The calculation is given below:

P(-2.33<z<1.50)=P(-2.33<z<0)+P(0<Z<-1.50)

=0.4901+0.4332=0.9233

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Most popular questions from this chapter

Independent random samples from normal populations produced the results shown in the next table.

Sample 1


Sample 2

1.23.11.72.83.0

4.22.73.63.9

a. Calculate the pooled estimate of σ2.

b. Do the data provide sufficient evidence to indicate that μ2&²µ³Ù;μ1? Test using α=.10.

c. Find a 90% confidence interval for (μ1−μ2).

d. Which of the two inferential procedures, the test of hypothesis in part b or the confidence interval in part c, provides more information about (μ1−μ2)?

Given the following values of x, s, and n, form a 90% confidence interval forσ2

a. x=21,s=2.5,n=50

b. x=1.3,s=0.02,n=15

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Drug content assessment. Refer to Exercise 8.16 (p. 467)and the Analytical Chemistry (Dec. 15, 2009) study in which scientists used high-performance liquid chromatography to determine the amount of drug in a tablet. Recall that 25 tablets were produced at each of two different, independent sites. The researchers want to determine if the two sites produced drug concentrations with different variances. A Minitab printout of the analysis follows. Locate the test statistic and p-value on the printout. Use these values α=.05and to conduct the appropriate test for the researchers.

Test and CI for two Variances: Content vs Site

Method

Null hypothesis α1α2=1

Alternative hypothesis α1α2≠1

F method was used. This method is accurate for normal data only.

Statistics

Site N St Dev Variance 95% CI for St Devs

1 25 3.067 9.406 (2.195,4.267)

2 25 3.339 11.147 (2.607,4.645)

Ratio of standard deviation =0.191

Ratio of variances=0.844

95% Confidence Intervals

Method CI for St Dev Ratio CI Variance Ratio

F (0.610, 1.384) (0.372, 1.915)

Tests

Method DF1 DF2 Test statistic p-value

F 24 24 0.84 0.681

A random sample of n observations is selected from a normal population to test the null hypothesis that σ2=25. Specify the rejection region for each of the following combinations of Ha,αand n.

a.Ha:σ2≠25;α=0.5;n=16

b.Ha:σ2>25;α=.10;n=15

c.Ha:σ2>25;α=.01;n=23

d. Ha:σ2<25;α=.01;n=13

e. Ha:σ2≠25;α=.10;n=7

f. Ha:σ2<25;α=.05;n=25

Studies have established that rudeness in the workplace can lead to retaliatory and counterproductive behaviour. However, there has been little research on how rude behaviours influence a victim’s task performance. Such a study was conducted, and the results were published in the Academy of Management Journal (Oct. 2007). College students enrolled in a management course were randomly assigned to two experimental conditions: rudeness condition (students) and control group (students). Each student was asked to write down as many uses for a brick as possible in minutes. For those students in the rudeness condition, the facilitator displayed rudeness by generally berating students for being irresponsible and unprofessional (due to a late-arriving confederate). No comments were made about the late-arriving confederate to students in the control group. The number of different uses for brick was recorded for each student and is shown below. Conduct a statistical analysis (at α=0.01) to determine if the true mean performance level for students in the rudeness condition is lower than the actual mean performance level for students in the control group.

The data is given below

Control Group:

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Rudeness Condition:

411181196511912757311191110789107114135478381591610071513921310

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