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a. For each situation, find Z (critical). Alpha ( α ) Form of the Test Z (critical) 0.05 One-tailed 0.10 Two-tailed 0.06 Two-tailed 0.01 One-tailed 0.02 Two-tailed b. For each situation, find the critical t score. Alpha ( α ) Form of the Test N t (critical) 0.10 Two-tailed 31 0.02 Two-tailed 24 0.01 Two-tailed 121 0.01 One-tailed 31 0.05 One-tailed 61 c. Compute the appropriate test statistic ( Z or t ) for each situations. Population Sample Z (obtained) or t (obtained) 1. μ = 2.40 σ = 0.75 X ¯ = 0.20 N = 200 2. μ = 17.1 X ¯ = 16.8 s = 0.9 N = 45 3. μ = 10.2 X ¯ = 9.4 s = 1.7 N = 150 4. P u = 0.57 P s = 0.60 N = 117 5. P u = 0.32 P s = 0.30 N = 322

BuyFind

Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836
BuyFind

Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836

Solutions

Chapter
Section
Chapter 7, Problem 7.1P
Textbook Problem
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a. For each situation, find Z(critical).

Alpha ( α ) Form of the Test Z(critical)
0.05 One-tailed
0.10 Two-tailed
0.06 Two-tailed
0.01 One-tailed
0.02 Two-tailed

b. For each situation, find the critical t score.

Alpha ( α ) Form of the Test N t(critical)
0.10 Two-tailed 31
0.02 Two-tailed 24
0.01 Two-tailed 121
0.01 One-tailed 31
0.05 One-tailed 61

c. Compute the appropriate test statistic (Z or t) for each situations.

Population Sample Z(obtained) or t(obtained)
1. μ = 2.40 σ = 0.75 X ¯ = 0.20 N = 200
2. μ = 17.1 X ¯ = 16.8 s = 0.9 N = 45
3. μ = 10.2 X ¯ = 9.4 s = 1.7 N = 150
4. P u = 0.57 P s = 0.60 N = 117
5. P u = 0.32 P s = 0.30 N = 322

Expert Solution
To determine

a)

To find:

The Z(critical) value for the given situation.

Answer to Problem 7.1P

Solution:

The Z(critical) values are given in Table (2).

Alpha (α) Form of the Test Z(critical)
0.05 One-tailed α=0.05 +1.645 or 1.645
0.10 Two-tailed α/2=0.05 ±1.645
0.06 Two-tailed α/2=0.03 ±1.881
0.01 One-tailed α=0.01 +2.326 or 2.326
0.02 Two-tailed α/2=0.01 ±2.326

Explanation of Solution

Given:

The given table of alpha values is,

Alpha (α) Form of the Test Z(critical)
0.05 One-tailed
0.10 Two-tailed
0.06 Two-tailed
0.01 One-tailed
0.02 Two-tailed

Description:

The critical region of a sampling distribution is the area that includes the unlikely sample outcomes.

The critical Z values are the areas under the standard normal curve.

For one-tailed critical values, the area is only on one side and the whole value of α is taken.

For two-tailed critical values, the area is equally divided on both the sides and the value of α taken is α/2.

Calculation:

The given table of alpha values is,

Alpha (α) Form of the Test Z(critical)
0.05 One-tailed
0.10 Two-tailed
0.06 Two-tailed
0.01 One-tailed
0.02 Two-tailed

Table (1)

The Z(critical) are obtained from the critical value table for standard normal variates.

For α=0.05, the one-tailed Z(critical) values are given from the critical value table as,

Z0.05=1.645

Or,

Z0.05=+1.645

Thus, for α=0.05, the one-tailed Z(critical) values are +1.645 or 1.645.

For α=0.10, the two-tailed Z(critical) values are obtained from the critical value table as,

Z0.10/2=Z0.05=1.645

And,

Z0.10/2=Z0.05=+1.645

Thus, for α=0.10, the two-tailed Z(critical) values are ±1.645.

For α=0.06, the two-tailed Z(critical) values are obtained from the critical value table as,

Z0.06/2=Z0.03=1.881

And,

Z0.06/2=Z0.03=+1.881

Thus, for α=0.06, the two-tailed Z(critical) values are ±1.881.

For α=0.01, the one-tailed Z(critical) values are obtained from the critical value table as,

Z0.01=2.326

Or,

Z0.01=+2.326

Thus, for α=0.01, the one-tailed Z(critical) values are +2.326 or 2.326.

For α=0.02, the two-tailed Z(critical) values are obtained from the critical value table as,

Z0.02/2=Z0.01=2.326

And,

Z0.02/2=Z0.01=+2.326

Thus, for α=0.02, the two-tailed Z(critical) values are ±2.326.

The obtained critical values are tabulated as,

Alpha (α) Form of the Test Z(critical)
0.05 One-tailed α=0.05 +1.645 or 1.645
0.10 Two-tailed α/2=0.05 ±1.645
0.06 Two-tailed α/2=0.03 ±1.881
0.01 One-tailed α=0.01 +2.326 or 2.326
0.02 Two-tailed α/2=0.01 ±2.326

Table (2)

Conclusion:

The Z(critical) values are given in Table (2).

Alpha (α) Form of the Test Z(critical)
0.05 One-tailed α=0.05 +1.645 or 1.645
0.10 Two-tailed α/2=0.05 ±1.645
0.06 Two-tailed α/2=0.03 ±1.881
0.01 One-tailed α=0.01 +2.326 or 2.326
0.02 Two-tailed α/2=0.01 ±2.326
Expert Solution
To determine

b)

To find:

The t(critical) value for the given situation.

Answer to Problem 7.1P

Solution:

The t(critical) values are given in Table (4).

Alpha (α) Form of the Test N t(critical)
0.10 Two-tailed α/2=0.05 31 ±1.697
0.02 Two-tailed α/2=0.01 24 ±2.5
0.01 Two-tailed α/2=0.005 121 ±2.617
0.01 One-tailed α=0.01 31 +2.457 or 2.457
0.05 One-tailed α=0.05 61 +1.671 or 1.671

Explanation of Solution

Given:

The given table of alpha values is,

Alpha (α) Form of the Test N t(critical)
0.10 Two-tailed 31
0.02 Two-tailed 24
0.01 Two-tailed 121
0.01 One-tailed 31
0.05 One-tailed 61

Description:

The critical region of a sampling distribution is the area that includes the unlikely sample outcomes.

The critical t values are the areas under the curve of t-distribution.

For one-tailed critical values, the area is only on one side and the whole value of α is taken.

For two-tailed critical values, the area is equally divided on both the sides and the value of α taken is α/2.

Calculation:

The given table of alpha values is,

Alpha (α) Form of the Test N t(critical)
0.10 Two-tailed 31
0.02 Two-tailed 24
0.01 Two-tailed 121
0.01 One-tailed 31
0.05 One-tailed 61

Table (3)

The t(critical) values are obtained from the t-distribution table with the given value of α and N1 degrees of freedom (df), where N is the sample size.s

For α=0.10 and N=31, the two-tailed t(critical) values are obtained from the critical value table as,

df=311=30

t(30,0.10/2)=t(30,0.05)=1.697

And,

t(30,0.10/2)=t(30,0.05)=+1.697

Thus, for α=0.10 and N=31, the two-tailed t(critical) values are ±1.697.

For α=0.02 and N=24, the two-tailed t(critical) values are obtained from the critical value table as,

df=241=23

t(23,0.02/2)=t(23,0.01)=2.5

And,

t(23,0.02/2)=t(23,0.01)=+2.5

Thus, for α=0.02 and N=24, the two-tailed t(critical) values are ±2.5.

For α=0.01 and N=121, the two-tailed t(critical) values are obtained from the critical value table as,

df=1211=120

t(120,0.01/2)=t(121,0.005)=2.617

And,

t(120,0.01/2)=t(121,0.005)=+2.617

Thus, for α=0.01 and N=121, the two-tailed t(critical) values are ±2.617.

For α=0.01 and N=31, the one-tailed t(critical) values are obtained from the critical value table as,

df=311=30

t(30,0.01)=2.457

Or,

t(30,0.01)=+2.457

Thus, for α=0.01 and N=31, the two-tailed t(critical) value is +2.457 or 2.457.

For α=0.05 and N=61, the one-tailed t(critical) values are obtained from the critical value table as,

df=611=60

t(60,0.05)=1.671

Or,

t(60,0.05)=+1.671

Thus, for α=0.05 and N=61, the two-tailed t(critical) value is +1.671 or 1.671.

The obtained critical values are tabulated as,

Alpha (α) Form of the Test N t(critical)
0.10 Two-tailed α/2=0.05 31 ±1.697
0.02 Two-tailed α/2=0.01 24 ±2.5
0.01 Two-tailed α/2=0.005 121 ±2.617
0.01 One-tailed α=0.01 31 +2.457 or 2.457
0.05 One-tailed α=0.05 61 +1.671 or 1.671

Table (4)

Conclusion:

The t(critical) values are given in Table (4).

Alpha (α) Form of the Test N t(critical)
0.10 Two-tailed α/2=0.05 31 ±1.697
0.02 Two-tailed α/2=0.01 24 ±2.5
0.01 Two-tailed α/2=0.005 121 ±2.617
0.01 One-tailed α=0.01 31 +2.457 or 2.457
0.05 One-tailed α=0.05 61 +1.671 or 1.671
Expert Solution
To determine

c)

To find:

The appropriate test statistics for the given situation.

Answer to Problem 7.1P

Solution:

The Z(critical) and t(critical) values are given in Table (6).

Population Sample Z(obtained) or t(obtained)
1. μ=2.40σ=0.75 X¯=0.20N=200 3.77
2. μ=17.1 X¯=16.8s=0.9N=45 2.24
3. μ=10.2 X¯=9.4s=1.7N=150 5.76
4. Pu=0.57 Ps=0.60N=117 0.66
5. Pu=0.32 Ps=0.30N=322 0.77

Explanation of Solution

Given:

The given table of information is,

Population Sample Z(obtained) or t(obtained)
1. μ=2.40σ=0.75 X¯=0.20N=200
2. μ=17.1 X¯=16.8s=0.9N=45
3. μ=10.2 X¯=9.4s=1.7N=150
4. Pu=0.57 Ps=0.60N=117
5. Pu=0.32 Ps=0.30N=322

Formula used:

For large samples with single mean and given standard deviation, the Z value is given by,

Z(obtained)=X¯μσ/N

Where, X¯ is the sample mean,

μ is the population,

σ is the population standard deviation and

N is the sample size.

For small samples with single sample mean and unknown standard deviation, the t value is given by,

t(obtained)=X¯μs/N1

Where, X¯ is the sample mean,

μ is the population mean,

s is the estimate population standard deviation and,

N is the sample size.

For large samples with single sample proportions, the Z value is given by,

Z(obtained)=PsPuPu(1Pu)/N

Where, Ps is the sample proportion,

Pu is the population proportion and,

N is the sample size.

Calculation:

The given table of alpha values is,

Population Sample Z(obtained) or t(obtained)
1. μ=2.40σ=0.75 X¯=0.20N=200
2. μ=17.1 X¯=16.8s=0.9N=45
3. μ=10.2 X¯=9.4s=1.7N=150
4. Pu=0.57 Ps=0.60N=117
5. Pu=0.32 Ps=0.30N=322

Table (5)

For large samples with single mean and given standard deviation, the Z value is given by,

Z(obtained)=X¯μσ/N ……(1)

For sample mean X¯=0.20, population mean μ=2.40, population standard deviation σ=0.75 and sample size N=200, the Z value is given by,

Substitute 2.20 for X¯, 2.40 for μ, 0.75 for σ and 200 for N in equation(1).

Z(obtained)=2.202.400.75/200=0.200.053=3.77

Thus, the obtained Z value is 3.77.

For small samples with single sample mean and unknown standard deviation, the t value is given by,

t(obtained)=X¯μs/N1 ……(2)

For sample mean X¯=16.8, population mean μ=17.1, sample standard deviation s=0.9 and sample size N=45, the t value is given by,

Substitute 16.8 for X¯, 17.1 for μ, 0.9 for s and 45 for N in equation(2).

t(obtained)=16.817.10.9/45=0.30.1342=2.24

Thus, the obtained t value is 2.24.

For sample mean X¯=9.4, population mean μ=10.2, sample standard deviation s=1.7 and sample size N=150, the t value is given by,

Substitute 9.4 for X¯, 10.2 for μ, 1.7 for s and 150 for N in equation(2).

t(obtained)=9.410.21.7/150=0.80.1388=5.76

Thus, the obtained t value is 5.76.

For large samples with single sample proportions, the Z value is given by,

Z(obtained)=PsPuPu(1Pu)/N ……(3)

For sample proportion Ps=0.60, population proportion Pu=0.57, and sample size N=117, the Z value is given by,

Substitute 0.60 for Ps, 0.57 for Pu and 117 for N in equation(3).

Z(obtained)=0.60.570.57(10.57)/117=0.030.0458=0.66

Thus, the obtained Z value is 0.66.

For sample proportion Ps=0.30, population proportion Pu=0.32, and sample size N=322, the Z value is given by,

Substitute 0.30 for Ps, 0.32 for Pu and 322 for N in equation(3).

Z(obtained)=0.300.320.32(10.32)/322=0.020.026=0.77

Thus, the obtained Z value is 0.77.

The obtained critical values are tabulated as,

Population Sample Z(obtained) or t(obtained)
1. μ=2.40σ=0.75 X¯=0.20N=200 3.77
2. μ=17.1 X¯=16.8s=0.9N=45 2.24
3. μ=10.2 X¯=9.4s=1.7N=150 5.76
4. Pu=0.57 Ps=0.60N=117 0.66
5. Pu=0.32 Ps=0.30N=322 0.77

Table (6)

Conclusion:

The Z(critical) and t(critical) values are given in Table (6).

Population Sample Z(obtained) or t(obtained)
1. μ=2.40σ=0.75 X¯=0.20N=200 3.77
2. μ=17.1 X¯=16.8s=0.9N=45 2.24
3. μ=10.2 X¯=9.4s=1.7N=150 5.76
4. Pu=0.57 Ps=0.60N=117 0.66
5. Pu=0.32 Ps=0.30N=322 0.77

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