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# Statistics: Critical Values (Statistics Project Sample)

Instructions:

The sample involves computation of the critical values and probabilities.

source..Content:

QUESTION 1

Mean,u=16.28

Standard deviation,Ïƒ=0.4

P(x<16) is given by

Wecompute the Z-value for the probability

For x=16, z= (16-16.28)/0.4=-0.7

P (z<-0.7) =

From the standard normal curve tables, we take the area to the left of 0.5 from the region z=0 to the region z=0.7.

P (z<-0.7) =0.5-P (00.7)

=0.5-0.2580=0.242

The probability that the average amount is between 16.20 and 16.30

P (16.2016.30)

We compute the Z-value for the 2 figures.

For x=16.20, Z= (16.20-16.28)/0.4=-0.2

For x=16.30, Z= (16.30-16.28)/0.4=0.05

P (-0.20.05)

From the standard normal table, we compute the area to the left side from z=-0.2 up to Z=0.This will be equal to Z=0 to Z=0.2

=P (-0.20.05) =0.0793+

The probability that the average amount is greater than 16.50 will computed as follows

We compute the Z-value of P(x>16.50)

Z-value= (16.50-16.28)/0.4=0.55

P (Z>0.55)

From the standard normal tables, we need to take the whole right side(0.5) and less the area from Z=0 to Z=0.55

P (Z>0.55) =0.5-0.2088=0.2912

The probability that the average amount is between 16.20 and 16.26

P (16.20

We need to compute the z-value for the above probability

For x=16.20, Z-value= (16.20-16.28)/0.4=-0.2

For x=16.26, Z-value = (16.26-16.28)/0.4=-0.05

P (-0.05

= -P (-0.2) +P (-0.05)

=- [1-P (Z<-0.2)] + [1-P (Z>-o.o5)]

=- [1-0.0793] + [1-0.0199]

= -0.9207+0.9801

=0.0594

99.9% of the time, the amount in the boxes will be 0.999*16.28=16.26 ounces

QUESTION 2

When it comes to apportioning students, you can either be a male or a female.So,the probabilityof student of any given gender being selected is 50%. The number of female students is 60 %( 0.6) while the number of male students is 40 %( 0.4)

Female, Q=0.6

Male, P=0.4

Students(male/female)

1/2

1/2

Number

0.6

0.4

The expected value,mean=1/2*6/10 +1/2*4/10

=0.3+0.2=0.5

The standard deviation for the random variable p is given by calculating the variance and then getting the square root.

Variance=X2ip-u2

=0.62*0.5 +0.42*0.5 -0.52

=0.18+0.08-0.25=0.01

Standard deviation=0.1

The probability that less than 58.5% of the students in the sample are female,P(á¿¤<0.585)

We compute the Z-value for the above probability.

Z-value= (0.585-0.5)/0.1

=0.085/0.1=0.85

P (á¿¤<0.85) =

From the standard normal table, we can easily get the above probability

P (á¿¤<0.85) =0.3023

The probability that more than 65% of the students are female is

P (á¿¤>0.65) =

We compute the Z-value of the probability that, more than 65% of the students are female.

Z-value= (0.65-0.5)/0.1=1.5

P (Z>1.5)

We can obtain the value for the above probability from the normal tables.

P (Z>1.5) =0.4332

The probability that the percentage of between 56% and 64% are female.

P (0.56<á¿¤<0.64).

The z-values are

For x=0.56, Z-value= (0.56-0.5)/0.1=0.6

Therefore, P (Z<0.6) =0.2257

For x=0.64, Z-value= (0.64-0.50)/0.1=1.4

Therefore,P (Z<1.4) =0.4192

The probability that the percentage of the students in the sample are female is =0.4192-0.2257=0.1935

The table below is the normal table used in solving question 1. The "0.1"s are running down, while the "0.01"s arerunning along.

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359

0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753

0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141

0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517

0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879

0.5 0.1915 0.195...

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