Find the level of significance. State the null and alternative hypothesis. Identify the tail of test.

Question

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Answer 1

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 8.1, Problem 24P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Identify the tail of test.

(a)

Expert Solution

Answer to Problem 24P

The level of significance is 0.01.

The null hypothesis is $H0:μ=28$ .

The alternative hypothesis is $H1:μ≠28$ .

The tail of the test is two-tailed.

Explanation of Solution

Calculation:

Let $μ$ denotes the Roger’s average red blood cell volume.

From the given information the value of $α$ is 0.01, and the data indicate that Roger’s red blood cell volume is different (either way) from $μ=28 ml/kg$ .

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H0:μ=28$

Alternative hypothesis:

$H1:μ≠28$

In this situation the alternative hypothesis is not equal indicates that the test is two-tailed test.

Hence, the tail of the test is two-tailed test.

(b)

To determine

Identify the sampling distribution to be used.

Explain how the sampling distribution is chosen.

Find the z value of the sample test statistic.

(b)

Expert Solution

Answer to Problem 24P

The sampling distribution to be used is $x¯$ distribution.

The $x¯$ sampling distribution is chosen because the x distribution is normal and population standard deviation is known.

The z value of the sample test statistic is 2.62.

Explanation of Solution

Calculation:

Conditions:

When the x distribution considered in the study has the normal distribution with the known population standard deviation $σ$ then the sampling distribution $x¯$ has normal distribution for any sample size n. The standardized z statistic is used for testing.
When the x distribution considered in the study is not normally distributed and the population standard deviation $σ$ is known then the sampling distribution $x¯$ has normal distribution if the sample size n is greater than or equal 30. That is, $n≥30$ .

Test statistic for z:

The z statistic value for sample test statistic $x¯$ is,

$z=x¯−μ(σn)$

In the formula $x¯$ is mean of a simple random sample, $μ$ is value stated in $H0$ , $σ$ is known standard deviation, and n is the sample size.

The distribution of x is assumed to be normal and the population standard deviation $σ=4.75$ . The x distribution considered in the study is normally distributed and population standard deviation is known. Hence, the sampling distribution to be used is $x¯$ distribution.

Z-statistic:

Substitute $x¯$ as 32.7, $μ$ as 28, $σ$ as 4.75, and n as 6 in the test statistic formula

$z=32.7−28(4.757)=4.71.7953=2.62$

Hence, the sample test statistic z is 2.62.

(c)

To determine

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

(c)

Expert Solution

Answer to Problem 24P

The P-value is 0.0088.

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tails, for the region of the curve to shade.
Enter the X value as 2.62.
Click OK.

Output using MINITAB software is given below:

Bundle: Understandable Statistics: Concepts And Methods, 12th + Webassign, Single-term Printed Access Card, Chapter 8.1, Problem 24P

From Minitab output, the P-value is 0.0044 which is one sided value.

The two-tailed P-value is,

$P-value=2×0.0044=0.0088$

Hence, the P-value of the test statistic is 0.0088.

(d)

To determine

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.01 or not.

(d)

Expert Solution

Answer to Problem 24P

The null hypothesis is rejected.

The data is statistically significant at level 0.01.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.0088.

Rejection rule:

If the P-value is less than or equal to $α$ , then reject the null hypothesis and the test is statistically significant. That is, $P-value≤α$ .

Conclusion:

The P-value is 0.0088 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, $0.0088(=P-value)<0.01(=α)$ .

By the rejection rule, the null hypothesis is rejected.

Hence, the data is statistically significant at level 0.01.

(e)

To determine

Interpret the conclusion in the context of the application.

(e)

Expert Solution

Explanation of Solution

Calculation:

From part (d), the null hypothesis is rejected. This shows that the researcher R’s average red blood cell volume is different (either way) from 28 ml/kg at 0.01 level of significance.

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Answer 2

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 8.1, Problem 24P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Identify the tail of test.

(a)

Expert Solution

Answer to Problem 24P

The level of significance is 0.01.

The null hypothesis is $H0:μ=28$ .

The alternative hypothesis is $H1:μ≠28$ .

The tail of the test is two-tailed.

Explanation of Solution

Calculation:

Let $μ$ denotes the Roger’s average red blood cell volume.

From the given information the value of $α$ is 0.01, and the data indicate that Roger’s red blood cell volume is different (either way) from $μ=28 ml/kg$ .

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H0:μ=28$

Alternative hypothesis:

$H1:μ≠28$

In this situation the alternative hypothesis is not equal indicates that the test is two-tailed test.

Hence, the tail of the test is two-tailed test.

(b)

To determine

Identify the sampling distribution to be used.

Explain how the sampling distribution is chosen.

Find the z value of the sample test statistic.

(b)

Expert Solution

Answer to Problem 24P

The sampling distribution to be used is $x¯$ distribution.

The $x¯$ sampling distribution is chosen because the x distribution is normal and population standard deviation is known.

The z value of the sample test statistic is 2.62.

Explanation of Solution

Calculation:

Conditions:

When the x distribution considered in the study has the normal distribution with the known population standard deviation $σ$ then the sampling distribution $x¯$ has normal distribution for any sample size n. The standardized z statistic is used for testing.
When the x distribution considered in the study is not normally distributed and the population standard deviation $σ$ is known then the sampling distribution $x¯$ has normal distribution if the sample size n is greater than or equal 30. That is, $n≥30$ .

Test statistic for z:

The z statistic value for sample test statistic $x¯$ is,

$z=x¯−μ(σn)$

In the formula $x¯$ is mean of a simple random sample, $μ$ is value stated in $H0$ , $σ$ is known standard deviation, and n is the sample size.

The distribution of x is assumed to be normal and the population standard deviation $σ=4.75$ . The x distribution considered in the study is normally distributed and population standard deviation is known. Hence, the sampling distribution to be used is $x¯$ distribution.

Z-statistic:

Substitute $x¯$ as 32.7, $μ$ as 28, $σ$ as 4.75, and n as 6 in the test statistic formula

$z=32.7−28(4.757)=4.71.7953=2.62$

Hence, the sample test statistic z is 2.62.

(c)

To determine

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

(c)

Expert Solution

Answer to Problem 24P

The P-value is 0.0088.

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tails, for the region of the curve to shade.
Enter the X value as 2.62.
Click OK.

Output using MINITAB software is given below:

Bundle: Understandable Statistics: Concepts And Methods, 12th + Webassign, Single-term Printed Access Card, Chapter 8.1, Problem 24P

From Minitab output, the P-value is 0.0044 which is one sided value.

The two-tailed P-value is,

$P-value=2×0.0044=0.0088$

Hence, the P-value of the test statistic is 0.0088.

(d)

To determine

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.01 or not.

(d)

Expert Solution

Answer to Problem 24P

The null hypothesis is rejected.

The data is statistically significant at level 0.01.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.0088.

Rejection rule:

If the P-value is less than or equal to $α$ , then reject the null hypothesis and the test is statistically significant. That is, $P-value≤α$ .

Conclusion:

The P-value is 0.0088 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, $0.0088(=P-value)<0.01(=α)$ .

By the rejection rule, the null hypothesis is rejected.

Hence, the data is statistically significant at level 0.01.

(e)

To determine

Interpret the conclusion in the context of the application.

(e)

Expert Solution

Explanation of Solution

Calculation:

From part (d), the null hypothesis is rejected. This shows that the researcher R’s average red blood cell volume is different (either way) from 28 ml/kg at 0.01 level of significance.

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Answer 3

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 8.1, Problem 24P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Identify the tail of test.

(a)

Expert Solution

Answer to Problem 24P

The level of significance is 0.01.

The null hypothesis is $H0:μ=28$ .

The alternative hypothesis is $H1:μ≠28$ .

The tail of the test is two-tailed.

Explanation of Solution

Calculation:

Let $μ$ denotes the Roger’s average red blood cell volume.

From the given information the value of $α$ is 0.01, and the data indicate that Roger’s red blood cell volume is different (either way) from $μ=28 ml/kg$ .

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H0:μ=28$

Alternative hypothesis:

$H1:μ≠28$

In this situation the alternative hypothesis is not equal indicates that the test is two-tailed test.

Hence, the tail of the test is two-tailed test.

(b)

To determine

Identify the sampling distribution to be used.

Explain how the sampling distribution is chosen.

Find the z value of the sample test statistic.

(b)

Expert Solution

Answer to Problem 24P

The sampling distribution to be used is $x¯$ distribution.

The $x¯$ sampling distribution is chosen because the x distribution is normal and population standard deviation is known.

The z value of the sample test statistic is 2.62.

Explanation of Solution

Calculation:

Conditions:

When the x distribution considered in the study has the normal distribution with the known population standard deviation $σ$ then the sampling distribution $x¯$ has normal distribution for any sample size n. The standardized z statistic is used for testing.
When the x distribution considered in the study is not normally distributed and the population standard deviation $σ$ is known then the sampling distribution $x¯$ has normal distribution if the sample size n is greater than or equal 30. That is, $n≥30$ .

Test statistic for z:

The z statistic value for sample test statistic $x¯$ is,

$z=x¯−μ(σn)$

In the formula $x¯$ is mean of a simple random sample, $μ$ is value stated in $H0$ , $σ$ is known standard deviation, and n is the sample size.

The distribution of x is assumed to be normal and the population standard deviation $σ=4.75$ . The x distribution considered in the study is normally distributed and population standard deviation is known. Hence, the sampling distribution to be used is $x¯$ distribution.

Z-statistic:

Substitute $x¯$ as 32.7, $μ$ as 28, $σ$ as 4.75, and n as 6 in the test statistic formula

$z=32.7−28(4.757)=4.71.7953=2.62$

Hence, the sample test statistic z is 2.62.

(c)

To determine

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

(c)

Expert Solution

Answer to Problem 24P

The P-value is 0.0088.

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tails, for the region of the curve to shade.
Enter the X value as 2.62.
Click OK.

Output using MINITAB software is given below:

Bundle: Understandable Statistics: Concepts And Methods, 12th + Webassign, Single-term Printed Access Card, Chapter 8.1, Problem 24P

From Minitab output, the P-value is 0.0044 which is one sided value.

The two-tailed P-value is,

$P-value=2×0.0044=0.0088$

Hence, the P-value of the test statistic is 0.0088.

(d)

To determine

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.01 or not.

(d)

Expert Solution

Answer to Problem 24P

The null hypothesis is rejected.

The data is statistically significant at level 0.01.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.0088.

Rejection rule:

If the P-value is less than or equal to $α$ , then reject the null hypothesis and the test is statistically significant. That is, $P-value≤α$ .

Conclusion:

The P-value is 0.0088 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, $0.0088(=P-value)<0.01(=α)$ .

By the rejection rule, the null hypothesis is rejected.

Hence, the data is statistically significant at level 0.01.

(e)

To determine

Interpret the conclusion in the context of the application.

(e)

Expert Solution

Explanation of Solution

Calculation:

From part (d), the null hypothesis is rejected. This shows that the researcher R’s average red blood cell volume is different (either way) from 28 ml/kg at 0.01 level of significance.

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Answer 4

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 8.1, Problem 24P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Identify the tail of test.

(a)

Expert Solution

Answer to Problem 24P

The level of significance is 0.01.

The null hypothesis is $H0:μ=28$ .

The alternative hypothesis is $H1:μ≠28$ .

The tail of the test is two-tailed.

Explanation of Solution

Calculation:

Let $μ$ denotes the Roger’s average red blood cell volume.

From the given information the value of $α$ is 0.01, and the data indicate that Roger’s red blood cell volume is different (either way) from $μ=28 ml/kg$ .

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H0:μ=28$

Alternative hypothesis:

$H1:μ≠28$

In this situation the alternative hypothesis is not equal indicates that the test is two-tailed test.

Hence, the tail of the test is two-tailed test.

(b)

To determine

Identify the sampling distribution to be used.

Explain how the sampling distribution is chosen.

Find the z value of the sample test statistic.

(b)

Expert Solution

Answer to Problem 24P

The sampling distribution to be used is $x¯$ distribution.

The $x¯$ sampling distribution is chosen because the x distribution is normal and population standard deviation is known.

The z value of the sample test statistic is 2.62.

Explanation of Solution

Calculation:

Conditions:

When the x distribution considered in the study has the normal distribution with the known population standard deviation $σ$ then the sampling distribution $x¯$ has normal distribution for any sample size n. The standardized z statistic is used for testing.
When the x distribution considered in the study is not normally distributed and the population standard deviation $σ$ is known then the sampling distribution $x¯$ has normal distribution if the sample size n is greater than or equal 30. That is, $n≥30$ .

Test statistic for z:

The z statistic value for sample test statistic $x¯$ is,

$z=x¯−μ(σn)$

In the formula $x¯$ is mean of a simple random sample, $μ$ is value stated in $H0$ , $σ$ is known standard deviation, and n is the sample size.

The distribution of x is assumed to be normal and the population standard deviation $σ=4.75$ . The x distribution considered in the study is normally distributed and population standard deviation is known. Hence, the sampling distribution to be used is $x¯$ distribution.

Z-statistic:

Substitute $x¯$ as 32.7, $μ$ as 28, $σ$ as 4.75, and n as 6 in the test statistic formula

$z=32.7−28(4.757)=4.71.7953=2.62$

Hence, the sample test statistic z is 2.62.

(c)

To determine

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

(c)

Expert Solution

Answer to Problem 24P

The P-value is 0.0088.

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tails, for the region of the curve to shade.
Enter the X value as 2.62.
Click OK.

Output using MINITAB software is given below:

Bundle: Understandable Statistics: Concepts And Methods, 12th + Webassign, Single-term Printed Access Card, Chapter 8.1, Problem 24P

From Minitab output, the P-value is 0.0044 which is one sided value.

The two-tailed P-value is,

$P-value=2×0.0044=0.0088$

Hence, the P-value of the test statistic is 0.0088.

(d)

To determine

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.01 or not.

(d)

Expert Solution

Answer to Problem 24P

The null hypothesis is rejected.

The data is statistically significant at level 0.01.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.0088.

Rejection rule:

If the P-value is less than or equal to $α$ , then reject the null hypothesis and the test is statistically significant. That is, $P-value≤α$ .

Conclusion:

The P-value is 0.0088 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, $0.0088(=P-value)<0.01(=α)$ .

By the rejection rule, the null hypothesis is rejected.

Hence, the data is statistically significant at level 0.01.

(e)

To determine

Interpret the conclusion in the context of the application.

(e)

Expert Solution

Explanation of Solution

Calculation:

From part (d), the null hypothesis is rejected. This shows that the researcher R’s average red blood cell volume is different (either way) from 28 ml/kg at 0.01 level of significance.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 8.1, Problem 24P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Identify the tail of test.

(a)

Expert Solution

Answer to Problem 24P

The level of significance is 0.01.

The null hypothesis is $H0:μ=28$ .

The alternative hypothesis is $H1:μ≠28$ .

The tail of the test is two-tailed.

Explanation of Solution

Calculation:

Let $μ$ denotes the Roger’s average red blood cell volume.

From the given information the value of $α$ is 0.01, and the data indicate that Roger’s red blood cell volume is different (either way) from $μ=28 ml/kg$ .

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H0:μ=28$

Alternative hypothesis:

$H1:μ≠28$

In this situation the alternative hypothesis is not equal indicates that the test is two-tailed test.

Hence, the tail of the test is two-tailed test.

(b)

To determine

Identify the sampling distribution to be used.

Explain how the sampling distribution is chosen.

Find the z value of the sample test statistic.

(b)

Expert Solution

Answer to Problem 24P

The sampling distribution to be used is $x¯$ distribution.

The $x¯$ sampling distribution is chosen because the x distribution is normal and population standard deviation is known.

The z value of the sample test statistic is 2.62.

Explanation of Solution

Calculation:

Conditions:

When the x distribution considered in the study has the normal distribution with the known population standard deviation $σ$ then the sampling distribution $x¯$ has normal distribution for any sample size n. The standardized z statistic is used for testing.
When the x distribution considered in the study is not normally distributed and the population standard deviation $σ$ is known then the sampling distribution $x¯$ has normal distribution if the sample size n is greater than or equal 30. That is, $n≥30$ .

Test statistic for z:

The z statistic value for sample test statistic $x¯$ is,

$z=x¯−μ(σn)$

In the formula $x¯$ is mean of a simple random sample, $μ$ is value stated in $H0$ , $σ$ is known standard deviation, and n is the sample size.

The distribution of x is assumed to be normal and the population standard deviation $σ=4.75$ . The x distribution considered in the study is normally distributed and population standard deviation is known. Hence, the sampling distribution to be used is $x¯$ distribution.

Z-statistic:

Substitute $x¯$ as 32.7, $μ$ as 28, $σ$ as 4.75, and n as 6 in the test statistic formula

$z=32.7−28(4.757)=4.71.7953=2.62$

Hence, the sample test statistic z is 2.62.

(c)

To determine

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

(c)

Expert Solution

Answer to Problem 24P

The P-value is 0.0088.

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tails, for the region of the curve to shade.
Enter the X value as 2.62.
Click OK.

Output using MINITAB software is given below:

Bundle: Understandable Statistics: Concepts And Methods, 12th + Webassign, Single-term Printed Access Card, Chapter 8.1, Problem 24P

From Minitab output, the P-value is 0.0044 which is one sided value.

The two-tailed P-value is,

$P-value=2×0.0044=0.0088$

Hence, the P-value of the test statistic is 0.0088.

(d)

To determine

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.01 or not.

(d)

Expert Solution

Answer to Problem 24P

The null hypothesis is rejected.

The data is statistically significant at level 0.01.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.0088.

Rejection rule:

If the P-value is less than or equal to $α$ , then reject the null hypothesis and the test is statistically significant. That is, $P-value≤α$ .

Conclusion:

The P-value is 0.0088 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, $0.0088(=P-value)<0.01(=α)$ .

By the rejection rule, the null hypothesis is rejected.

Hence, the data is statistically significant at level 0.01.

(e)

To determine

Interpret the conclusion in the context of the application.

(e)

Expert Solution

Explanation of Solution

Calculation:

From part (d), the null hypothesis is rejected. This shows that the researcher R’s average red blood cell volume is different (either way) from 28 ml/kg at 0.01 level of significance.

Answer 6

Chapter 8.1, Problem 24P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Identify the tail of test.

(a)

Expert Solution

Answer to Problem 24P

The level of significance is 0.01.

The null hypothesis is $H0:μ=28$ .

The alternative hypothesis is $H1:μ≠28$ .

The tail of the test is two-tailed.

Explanation of Solution

Calculation:

Let $μ$ denotes the Roger’s average red blood cell volume.

From the given information the value of $α$ is 0.01, and the data indicate that Roger’s red blood cell volume is different (either way) from $μ=28 ml/kg$ .

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H0:μ=28$

Alternative hypothesis:

$H1:μ≠28$

In this situation the alternative hypothesis is not equal indicates that the test is two-tailed test.

Hence, the tail of the test is two-tailed test.

Answer 7

Chapter 8.1, Problem 24P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Identify the tail of test.

Answer 8

(a)

Expert Solution

Answer to Problem 24P

The level of significance is 0.01.

The null hypothesis is $H0:μ=28$ .

The alternative hypothesis is $H1:μ≠28$ .

The tail of the test is two-tailed.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

Let $μ$ denotes the Roger’s average red blood cell volume.

From the given information the value of $α$ is 0.01, and the data indicate that Roger’s red blood cell volume is different (either way) from $μ=28 ml/kg$ .

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H0:μ=28$

Alternative hypothesis:

$H1:μ≠28$

In this situation the alternative hypothesis is not equal indicates that the test is two-tailed test.

Hence, the tail of the test is two-tailed test.

Answer 13

(b)

To determine

Identify the sampling distribution to be used.

Explain how the sampling distribution is chosen.

Find the z value of the sample test statistic.

(b)

Expert Solution

Answer to Problem 24P

The sampling distribution to be used is $x¯$ distribution.

The $x¯$ sampling distribution is chosen because the x distribution is normal and population standard deviation is known.

The z value of the sample test statistic is 2.62.

Explanation of Solution

Calculation:

Conditions:

When the x distribution considered in the study has the normal distribution with the known population standard deviation $σ$ then the sampling distribution $x¯$ has normal distribution for any sample size n. The standardized z statistic is used for testing.
When the x distribution considered in the study is not normally distributed and the population standard deviation $σ$ is known then the sampling distribution $x¯$ has normal distribution if the sample size n is greater than or equal 30. That is, $n≥30$ .

Test statistic for z:

The z statistic value for sample test statistic $x¯$ is,

$z=x¯−μ(σn)$

In the formula $x¯$ is mean of a simple random sample, $μ$ is value stated in $H0$ , $σ$ is known standard deviation, and n is the sample size.

The distribution of x is assumed to be normal and the population standard deviation $σ=4.75$ . The x distribution considered in the study is normally distributed and population standard deviation is known. Hence, the sampling distribution to be used is $x¯$ distribution.

Z-statistic:

Substitute $x¯$ as 32.7, $μ$ as 28, $σ$ as 4.75, and n as 6 in the test statistic formula

$z=32.7−28(4.757)=4.71.7953=2.62$

Hence, the sample test statistic z is 2.62.

Answer 14

(b)

To determine

Identify the sampling distribution to be used.

Explain how the sampling distribution is chosen.

Find the z value of the sample test statistic.

Answer 15

(b)

Expert Solution

Answer to Problem 24P

The sampling distribution to be used is $x¯$ distribution.

The $x¯$ sampling distribution is chosen because the x distribution is normal and population standard deviation is known.

The z value of the sample test statistic is 2.62.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

Conditions:

When the x distribution considered in the study has the normal distribution with the known population standard deviation $σ$ then the sampling distribution $x¯$ has normal distribution for any sample size n. The standardized z statistic is used for testing.
When the x distribution considered in the study is not normally distributed and the population standard deviation $σ$ is known then the sampling distribution $x¯$ has normal distribution if the sample size n is greater than or equal 30. That is, $n≥30$ .

Test statistic for z:

The z statistic value for sample test statistic $x¯$ is,

$z=x¯−μ(σn)$

In the formula $x¯$ is mean of a simple random sample, $μ$ is value stated in $H0$ , $σ$ is known standard deviation, and n is the sample size.

The distribution of x is assumed to be normal and the population standard deviation $σ=4.75$ . The x distribution considered in the study is normally distributed and population standard deviation is known. Hence, the sampling distribution to be used is $x¯$ distribution.

Z-statistic:

Substitute $x¯$ as 32.7, $μ$ as 28, $σ$ as 4.75, and n as 6 in the test statistic formula

$z=32.7−28(4.757)=4.71.7953=2.62$

Hence, the sample test statistic z is 2.62.

Answer 20

(c)

To determine

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

(c)

Expert Solution

Answer to Problem 24P

The P-value is 0.0088.

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tails, for the region of the curve to shade.
Enter the X value as 2.62.
Click OK.

Output using MINITAB software is given below:

Bundle: Understandable Statistics: Concepts And Methods, 12th + Webassign, Single-term Printed Access Card, Chapter 8.1, Problem 24P

From Minitab output, the P-value is 0.0044 which is one sided value.

The two-tailed P-value is,

$P-value=2×0.0044=0.0088$

Hence, the P-value of the test statistic is 0.0088.

Answer 21

(c)

To determine

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

Answer 22

(c)

Expert Solution

Answer to Problem 24P

The P-value is 0.0088.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

Choose Graph > Probability Distribution Plot choose View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tails, for the region of the curve to shade.
Enter the X value as 2.62.
Click OK.

Output using MINITAB software is given below:

Bundle: Understandable Statistics: Concepts And Methods, 12th + Webassign, Single-term Printed Access Card, Chapter 8.1, Problem 24P

From Minitab output, the P-value is 0.0044 which is one sided value.

The two-tailed P-value is,

$P-value=2×0.0044=0.0088$

Hence, the P-value of the test statistic is 0.0088.

Answer 27

(d)

To determine

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.01 or not.

(d)

Expert Solution

Answer to Problem 24P

The null hypothesis is rejected.

The data is statistically significant at level 0.01.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.0088.

Rejection rule:

If the P-value is less than or equal to $α$ , then reject the null hypothesis and the test is statistically significant. That is, $P-value≤α$ .

Conclusion:

The P-value is 0.0088 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, $0.0088(=P-value)<0.01(=α)$ .

By the rejection rule, the null hypothesis is rejected.

Hence, the data is statistically significant at level 0.01.

Answer 28

(d)

To determine

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.01 or not.

Answer 29

(d)

Expert Solution

Answer to Problem 24P

The null hypothesis is rejected.

The data is statistically significant at level 0.01.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation:

From part (c), the P-value is 0.0088.

Rejection rule:

If the P-value is less than or equal to $α$ , then reject the null hypothesis and the test is statistically significant. That is, $P-value≤α$ .

Conclusion:

The P-value is 0.0088 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, $0.0088(=P-value)<0.01(=α)$ .

By the rejection rule, the null hypothesis is rejected.

Hence, the data is statistically significant at level 0.01.

Answer 34

(e)

To determine

Interpret the conclusion in the context of the application.

(e)

Expert Solution

Explanation of Solution

Calculation:

From part (d), the null hypothesis is rejected. This shows that the researcher R’s average red blood cell volume is different (either way) from 28 ml/kg at 0.01 level of significance.

Answer 35

(e)

To determine

Interpret the conclusion in the context of the application.

Answer 36

(e)

Expert Solution

Explanation of Solution

Calculation:

From part (d), the null hypothesis is rejected. This shows that the researcher R’s average red blood cell volume is different (either way) from 28 ml/kg at 0.01 level of significance.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Find the level of significance. State the null and alternative hypothesis. Identify the tail of test.

Videos

Find the level of significance.

State the null and alternative hypothesis.

Identify the tail of test.

Answer to Problem 24P

Explanation of Solution

Identify the sampling distribution to be used.

Explain how the sampling distribution is chosen.

Find the z value of the sample test statistic.

Answer to Problem 24P

Explanation of Solution

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

Answer to Problem 24P

Explanation of Solution

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.01 or not.

Answer to Problem 24P

Explanation of Solution

Interpret the conclusion in the context of the application.

Explanation of Solution

Want to see more full solutions like this?

Chapter 8 Solutions

Find the level of significance. State the null and alternative hypothesis. Identify the tail of test.

Videos

Find the level of significance. State the null and alternative hypothesis. Identify the tail of test.

Answer to Problem 24P

Explanation of Solution

Identify the sampling distribution to be used. Explain how the sampling distribution is chosen. Find the z value of the sample test statistic.

Answer to Problem 24P

Explanation of Solution

Find the P-value. Draw the sampling distribution by showing the area corresponding to the P-value.

Answer to Problem 24P

Explanation of Solution

Check whether the null hypothesis is rejecting or fail to reject. Identify whether the data statistically significant at level 0.01 or not.

Answer to Problem 24P

Explanation of Solution

Interpret the conclusion in the context of the application.

Explanation of Solution

Want to see more full solutions like this?

Chapter 8 Solutions

Find the level of significance.

State the null and alternative hypothesis.

Identify the tail of test.

Identify the sampling distribution to be used.

Explain how the sampling distribution is chosen.

Find the z value of the sample test statistic.

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.01 or not.