Students at the Akademia Podlaka conducted an experiment to determine whether the Belgium-minted Euro coin was equally likely to land heads up or tails up. Coins were spun on a smooth surface, and in 250 spins, 140 landed with the heads side up ( New Scientist , January 4, 2002). a. Should the students interpret this result as convincing evidence that the proportion of the time the coin would land heads up is not 0.5? Test the relevant hypotheses using α = 0.01. b. Would your conclusion be different if a significance level of 0.05 had been used? Explain.

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Answer 1

Textbook Question

Chapter 10, Problem 88CR

Students at the Akademia Podlaka conducted an experiment to determine whether the Belgium-minted Euro coin was equally likely to land heads up or tails up. Coins were spun on a smooth surface, and in 250 spins, 140 landed with the heads side up (New Scientist, January 4, 2002).

a. Should the students interpret this result as convincing evidence that the proportion of the time the coin would land heads up is not 0.5? Test the relevant hypotheses using α = 0.01.
b. Would your conclusion be different if a significance level of 0.05 had been used? Explain.

a.

Expert Solution

To determine

Check whether the sample data provide convincing evidence that the proportion of time a coin landed as heads up is different from 0.5 at 0.01 significance level.

Answer to Problem 88CR

No, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

Explanation of Solution

Calculation:

In a sample of 250 spinning of Belgium-minted Euro coin, 140 spins were founded to turn up as heads.

Step 1:

Population characteristic of interest:

Let p be the population characteristic of interest.

In this context, p denotes the population proportion of spins that landed as heads up.

Step 2:

Null hypothesis: $H0:p=0.5$

That is, the proportion of spins that landed as heads up is 0.5.

Step 3:

Alternative hypothesis: $Ha:p≠0.5$

That is, the proportion of spins that landed as heads up is different from 0.5.

Step 4:

Significance level, $α$ :

It is given that the significance level, $α=0.01$ .

Step 5:

Test statistic, z:

$z=p^−pp(1−p)n$ ,

Where, $p^$ be the sample proportion, p is the hypothesized proportion, and n is the sample size.

Substitute the hypothesized proportion, $p=0.5$ , in the test statistic.

$z=p^−0.50.5(1−0.5)n$

Here, the sample proportion $p^$ is not known.

Step 6:

Assumptions:

Let $p^$ be the sample proportion from a random sample.
The large sample z test can be used if the sample size n satisfies the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .
The sample size should not be greater than 10% of the population size.

Requirement check:

It is assumed that the sample is from a random sample.
Check the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .

$n(hypothesized value)=np=250(0.5)=175>10$

$n(1−hypothesized value)=n(1−p)=250(1−0.5)=250(0.5)=175>10$

Since $n(hypothesized value)$ and $n(1−hypothesized value)$ are greater than 10, the sample size requirement is met.

Therefore, the large sample z test is appropriate.

Although, the population size is not known, it is reasonable to assume that the sample size of 250 acts as the representative for all the spins of Euro coins. It is also definite that the sample size is less than the 10% of the population of the spins of Euro coins.

Step 7:

The value of the test statistic is obtained as follows:

$z=p^−0.50.5(1−0.5)n$

The sample proportion, $p^$ , is obtained as follows:

$p^=xn$

Where, x is the number of observations of interest and n is the sample size.

Substitute the corresponding values to get the sample proportion.

$p^=140250=0.56$

By substituting the value of sample proportion, the test statistic is obtained as follows:

$z=0.56−0.50.5(1−0.5)250=0.060.032≈1.90$

Thus, the value of test statistic is 1.90.

Step 8:

P-value:

In this context, the alternative hypothesis denotes that the test carried out is two-tailed test. Therefore, the P-value is the area under the z curve and the two-tailed of the calculated z value.

The P-value for the test statistic value of 1.90 is obtained as follows:

$P-value=2×{1−Area to the Right of 1.90}=2×{1−P(z<1.90)}$

Use standard normal probabilities (Cumulative z curve areas) table to find the z-value.

Procedure:

For z at 1.90:

 Locate 1.9 in the left column of the table.

 Obtain the value in the corresponding row below .00.

That is, $P(z<1.90)=0.9713$ .

The probability of the event is as given below:

$P-value=2{1−P(z<1.90)}=2{1−0.9713}=2{0.0287}=0.0574$

Thus, the P-value for the test statistic of 1.90 is 0.0574.

Step 9:

Decision rule:

If $P-value≤Significance level$ , then reject the null hypothesis $H0$ .

If $P-value>Significance level$ , then fail to reject the null hypothesis $H0$ .

Here, the $P-value$ of 0.0574 is greater than the significance level 0.01.

That is, $P-value(=0.0574)>Significance level(=0.01)$ .

The decision is that the null hypothesis is not rejected.

Conclusion:

Hence, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

b.

Expert Solution

To determine

Explain whether the conclusion for the hypothesis test is different at significance level $α=0.05$ .

Explanation of Solution

Increasing significance level increases, the chance to reject the null hypothesis.

It is given that the significance level $α=0.05$ .

Here, the $P-value$ of 0.0574 is less than the significance level 0.05.

That is, $P-value(=0.0574)> Significance level(=0.05)$ .

Therefore, it can be concluded that the null hypothesis is not rejected at the 0.05 significance level.

Hence, there is no evidence that the proportion of time the coin landed as heads up is different from 0.5.

In this context, even if the significance is increased, there is no change in the conclusion.

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The following is a list of data on the duration of a sample of 200 outbreaks, in hours. 107 73 68 97 76 79 94 59 98 57 54 65 71 70 84 88 62 82 61 79 98 66 62 79 86 68 74 61 62 116 65 88 64 79 78 74 92 75 5289 85 28 73 80 68 78 89 72 78 88 77 103 88 63 68 90 62 89 71 71 74 222 R 82 79 70 ST☑ 65 98 77 86 58 69 88 81 74 70 65 81 75 81 78 90 78 96 75 KRRE F S 62 94 62 79 83 93 135 71 85 84 83 63 61 65 83 70 70 81 77 72 84 33 62 92 65 67 59 58 66 66 94 77 63 71 101 78 43 78 66 75 68 76 59 67 61 71 64 76 72 77 74 65 82 86 66 86 68 85 27% 96 72 77 60 67 87 83 68 72 74 91 76 83 งงง 8 སྐྱ ཐྭ ༄ ཏྱཾ 89 81 71 85 99 59 92 87 84 75 77 51 45 80 84 93 69 76 89 75 67 92 89 82 96 77 102 66 68 61 73 72 76 73 77 79 94 63 59 62 71 81 65 73 63 63 89 82 64 85 92 64 73 a. What is the variable? What type? b. Construct an interval-frequency table, with columns containing: class mark, absolute frequency, relative frequency, cumulative frequency, cumulative relative frequency, and percentage frequency.

Answer 2

Textbook Question

Chapter 10, Problem 88CR

Students at the Akademia Podlaka conducted an experiment to determine whether the Belgium-minted Euro coin was equally likely to land heads up or tails up. Coins were spun on a smooth surface, and in 250 spins, 140 landed with the heads side up (New Scientist, January 4, 2002).

a. Should the students interpret this result as convincing evidence that the proportion of the time the coin would land heads up is not 0.5? Test the relevant hypotheses using α = 0.01.
b. Would your conclusion be different if a significance level of 0.05 had been used? Explain.

a.

Expert Solution

To determine

Check whether the sample data provide convincing evidence that the proportion of time a coin landed as heads up is different from 0.5 at 0.01 significance level.

Answer to Problem 88CR

No, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

Explanation of Solution

Calculation:

In a sample of 250 spinning of Belgium-minted Euro coin, 140 spins were founded to turn up as heads.

Step 1:

Population characteristic of interest:

Let p be the population characteristic of interest.

In this context, p denotes the population proportion of spins that landed as heads up.

Step 2:

Null hypothesis: $H0:p=0.5$

That is, the proportion of spins that landed as heads up is 0.5.

Step 3:

Alternative hypothesis: $Ha:p≠0.5$

That is, the proportion of spins that landed as heads up is different from 0.5.

Step 4:

Significance level, $α$ :

It is given that the significance level, $α=0.01$ .

Step 5:

Test statistic, z:

$z=p^−pp(1−p)n$ ,

Where, $p^$ be the sample proportion, p is the hypothesized proportion, and n is the sample size.

Substitute the hypothesized proportion, $p=0.5$ , in the test statistic.

$z=p^−0.50.5(1−0.5)n$

Here, the sample proportion $p^$ is not known.

Step 6:

Assumptions:

Let $p^$ be the sample proportion from a random sample.
The large sample z test can be used if the sample size n satisfies the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .
The sample size should not be greater than 10% of the population size.

Requirement check:

It is assumed that the sample is from a random sample.
Check the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .

$n(hypothesized value)=np=250(0.5)=175>10$

$n(1−hypothesized value)=n(1−p)=250(1−0.5)=250(0.5)=175>10$

Since $n(hypothesized value)$ and $n(1−hypothesized value)$ are greater than 10, the sample size requirement is met.

Therefore, the large sample z test is appropriate.

Although, the population size is not known, it is reasonable to assume that the sample size of 250 acts as the representative for all the spins of Euro coins. It is also definite that the sample size is less than the 10% of the population of the spins of Euro coins.

Step 7:

The value of the test statistic is obtained as follows:

$z=p^−0.50.5(1−0.5)n$

The sample proportion, $p^$ , is obtained as follows:

$p^=xn$

Where, x is the number of observations of interest and n is the sample size.

Substitute the corresponding values to get the sample proportion.

$p^=140250=0.56$

By substituting the value of sample proportion, the test statistic is obtained as follows:

$z=0.56−0.50.5(1−0.5)250=0.060.032≈1.90$

Thus, the value of test statistic is 1.90.

Step 8:

P-value:

In this context, the alternative hypothesis denotes that the test carried out is two-tailed test. Therefore, the P-value is the area under the z curve and the two-tailed of the calculated z value.

The P-value for the test statistic value of 1.90 is obtained as follows:

$P-value=2×{1−Area to the Right of 1.90}=2×{1−P(z<1.90)}$

Use standard normal probabilities (Cumulative z curve areas) table to find the z-value.

Procedure:

For z at 1.90:

 Locate 1.9 in the left column of the table.

 Obtain the value in the corresponding row below .00.

That is, $P(z<1.90)=0.9713$ .

The probability of the event is as given below:

$P-value=2{1−P(z<1.90)}=2{1−0.9713}=2{0.0287}=0.0574$

Thus, the P-value for the test statistic of 1.90 is 0.0574.

Step 9:

Decision rule:

If $P-value≤Significance level$ , then reject the null hypothesis $H0$ .

If $P-value>Significance level$ , then fail to reject the null hypothesis $H0$ .

Here, the $P-value$ of 0.0574 is greater than the significance level 0.01.

That is, $P-value(=0.0574)>Significance level(=0.01)$ .

The decision is that the null hypothesis is not rejected.

Conclusion:

Hence, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

b.

Expert Solution

To determine

Explain whether the conclusion for the hypothesis test is different at significance level $α=0.05$ .

Explanation of Solution

Increasing significance level increases, the chance to reject the null hypothesis.

It is given that the significance level $α=0.05$ .

Here, the $P-value$ of 0.0574 is less than the significance level 0.05.

That is, $P-value(=0.0574)> Significance level(=0.05)$ .

Therefore, it can be concluded that the null hypothesis is not rejected at the 0.05 significance level.

Hence, there is no evidence that the proportion of time the coin landed as heads up is different from 0.5.

In this context, even if the significance is increased, there is no change in the conclusion.

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Answer 3

Textbook Question

Chapter 10, Problem 88CR

Students at the Akademia Podlaka conducted an experiment to determine whether the Belgium-minted Euro coin was equally likely to land heads up or tails up. Coins were spun on a smooth surface, and in 250 spins, 140 landed with the heads side up (New Scientist, January 4, 2002).

a. Should the students interpret this result as convincing evidence that the proportion of the time the coin would land heads up is not 0.5? Test the relevant hypotheses using α = 0.01.
b. Would your conclusion be different if a significance level of 0.05 had been used? Explain.

a.

Expert Solution

To determine

Check whether the sample data provide convincing evidence that the proportion of time a coin landed as heads up is different from 0.5 at 0.01 significance level.

Answer to Problem 88CR

No, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

Explanation of Solution

Calculation:

In a sample of 250 spinning of Belgium-minted Euro coin, 140 spins were founded to turn up as heads.

Step 1:

Population characteristic of interest:

Let p be the population characteristic of interest.

In this context, p denotes the population proportion of spins that landed as heads up.

Step 2:

Null hypothesis: $H0:p=0.5$

That is, the proportion of spins that landed as heads up is 0.5.

Step 3:

Alternative hypothesis: $Ha:p≠0.5$

That is, the proportion of spins that landed as heads up is different from 0.5.

Step 4:

Significance level, $α$ :

It is given that the significance level, $α=0.01$ .

Step 5:

Test statistic, z:

$z=p^−pp(1−p)n$ ,

Where, $p^$ be the sample proportion, p is the hypothesized proportion, and n is the sample size.

Substitute the hypothesized proportion, $p=0.5$ , in the test statistic.

$z=p^−0.50.5(1−0.5)n$

Here, the sample proportion $p^$ is not known.

Step 6:

Assumptions:

Let $p^$ be the sample proportion from a random sample.
The large sample z test can be used if the sample size n satisfies the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .
The sample size should not be greater than 10% of the population size.

Requirement check:

It is assumed that the sample is from a random sample.
Check the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .

$n(hypothesized value)=np=250(0.5)=175>10$

$n(1−hypothesized value)=n(1−p)=250(1−0.5)=250(0.5)=175>10$

Since $n(hypothesized value)$ and $n(1−hypothesized value)$ are greater than 10, the sample size requirement is met.

Therefore, the large sample z test is appropriate.

Although, the population size is not known, it is reasonable to assume that the sample size of 250 acts as the representative for all the spins of Euro coins. It is also definite that the sample size is less than the 10% of the population of the spins of Euro coins.

Step 7:

The value of the test statistic is obtained as follows:

$z=p^−0.50.5(1−0.5)n$

The sample proportion, $p^$ , is obtained as follows:

$p^=xn$

Where, x is the number of observations of interest and n is the sample size.

Substitute the corresponding values to get the sample proportion.

$p^=140250=0.56$

By substituting the value of sample proportion, the test statistic is obtained as follows:

$z=0.56−0.50.5(1−0.5)250=0.060.032≈1.90$

Thus, the value of test statistic is 1.90.

Step 8:

P-value:

In this context, the alternative hypothesis denotes that the test carried out is two-tailed test. Therefore, the P-value is the area under the z curve and the two-tailed of the calculated z value.

The P-value for the test statistic value of 1.90 is obtained as follows:

$P-value=2×{1−Area to the Right of 1.90}=2×{1−P(z<1.90)}$

Use standard normal probabilities (Cumulative z curve areas) table to find the z-value.

Procedure:

For z at 1.90:

 Locate 1.9 in the left column of the table.

 Obtain the value in the corresponding row below .00.

That is, $P(z<1.90)=0.9713$ .

The probability of the event is as given below:

$P-value=2{1−P(z<1.90)}=2{1−0.9713}=2{0.0287}=0.0574$

Thus, the P-value for the test statistic of 1.90 is 0.0574.

Step 9:

Decision rule:

If $P-value≤Significance level$ , then reject the null hypothesis $H0$ .

If $P-value>Significance level$ , then fail to reject the null hypothesis $H0$ .

Here, the $P-value$ of 0.0574 is greater than the significance level 0.01.

That is, $P-value(=0.0574)>Significance level(=0.01)$ .

The decision is that the null hypothesis is not rejected.

Conclusion:

Hence, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

b.

Expert Solution

To determine

Explain whether the conclusion for the hypothesis test is different at significance level $α=0.05$ .

Explanation of Solution

Increasing significance level increases, the chance to reject the null hypothesis.

It is given that the significance level $α=0.05$ .

Here, the $P-value$ of 0.0574 is less than the significance level 0.05.

That is, $P-value(=0.0574)> Significance level(=0.05)$ .

Therefore, it can be concluded that the null hypothesis is not rejected at the 0.05 significance level.

Hence, there is no evidence that the proportion of time the coin landed as heads up is different from 0.5.

In this context, even if the significance is increased, there is no change in the conclusion.

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Answer 4

Textbook Question

Chapter 10, Problem 88CR

Students at the Akademia Podlaka conducted an experiment to determine whether the Belgium-minted Euro coin was equally likely to land heads up or tails up. Coins were spun on a smooth surface, and in 250 spins, 140 landed with the heads side up (New Scientist, January 4, 2002).

a. Should the students interpret this result as convincing evidence that the proportion of the time the coin would land heads up is not 0.5? Test the relevant hypotheses using α = 0.01.
b. Would your conclusion be different if a significance level of 0.05 had been used? Explain.

a.

Expert Solution

To determine

Check whether the sample data provide convincing evidence that the proportion of time a coin landed as heads up is different from 0.5 at 0.01 significance level.

Answer to Problem 88CR

No, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

Explanation of Solution

Calculation:

In a sample of 250 spinning of Belgium-minted Euro coin, 140 spins were founded to turn up as heads.

Step 1:

Population characteristic of interest:

Let p be the population characteristic of interest.

In this context, p denotes the population proportion of spins that landed as heads up.

Step 2:

Null hypothesis: $H0:p=0.5$

That is, the proportion of spins that landed as heads up is 0.5.

Step 3:

Alternative hypothesis: $Ha:p≠0.5$

That is, the proportion of spins that landed as heads up is different from 0.5.

Step 4:

Significance level, $α$ :

It is given that the significance level, $α=0.01$ .

Step 5:

Test statistic, z:

$z=p^−pp(1−p)n$ ,

Where, $p^$ be the sample proportion, p is the hypothesized proportion, and n is the sample size.

Substitute the hypothesized proportion, $p=0.5$ , in the test statistic.

$z=p^−0.50.5(1−0.5)n$

Here, the sample proportion $p^$ is not known.

Step 6:

Assumptions:

Let $p^$ be the sample proportion from a random sample.
The large sample z test can be used if the sample size n satisfies the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .
The sample size should not be greater than 10% of the population size.

Requirement check:

It is assumed that the sample is from a random sample.
Check the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .

$n(hypothesized value)=np=250(0.5)=175>10$

$n(1−hypothesized value)=n(1−p)=250(1−0.5)=250(0.5)=175>10$

Since $n(hypothesized value)$ and $n(1−hypothesized value)$ are greater than 10, the sample size requirement is met.

Therefore, the large sample z test is appropriate.

Although, the population size is not known, it is reasonable to assume that the sample size of 250 acts as the representative for all the spins of Euro coins. It is also definite that the sample size is less than the 10% of the population of the spins of Euro coins.

Step 7:

The value of the test statistic is obtained as follows:

$z=p^−0.50.5(1−0.5)n$

The sample proportion, $p^$ , is obtained as follows:

$p^=xn$

Where, x is the number of observations of interest and n is the sample size.

Substitute the corresponding values to get the sample proportion.

$p^=140250=0.56$

By substituting the value of sample proportion, the test statistic is obtained as follows:

$z=0.56−0.50.5(1−0.5)250=0.060.032≈1.90$

Thus, the value of test statistic is 1.90.

Step 8:

P-value:

In this context, the alternative hypothesis denotes that the test carried out is two-tailed test. Therefore, the P-value is the area under the z curve and the two-tailed of the calculated z value.

The P-value for the test statistic value of 1.90 is obtained as follows:

$P-value=2×{1−Area to the Right of 1.90}=2×{1−P(z<1.90)}$

Use standard normal probabilities (Cumulative z curve areas) table to find the z-value.

Procedure:

For z at 1.90:

 Locate 1.9 in the left column of the table.

 Obtain the value in the corresponding row below .00.

That is, $P(z<1.90)=0.9713$ .

The probability of the event is as given below:

$P-value=2{1−P(z<1.90)}=2{1−0.9713}=2{0.0287}=0.0574$

Thus, the P-value for the test statistic of 1.90 is 0.0574.

Step 9:

Decision rule:

If $P-value≤Significance level$ , then reject the null hypothesis $H0$ .

If $P-value>Significance level$ , then fail to reject the null hypothesis $H0$ .

Here, the $P-value$ of 0.0574 is greater than the significance level 0.01.

That is, $P-value(=0.0574)>Significance level(=0.01)$ .

The decision is that the null hypothesis is not rejected.

Conclusion:

Hence, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

b.

Expert Solution

To determine

Explain whether the conclusion for the hypothesis test is different at significance level $α=0.05$ .

Explanation of Solution

Increasing significance level increases, the chance to reject the null hypothesis.

It is given that the significance level $α=0.05$ .

Here, the $P-value$ of 0.0574 is less than the significance level 0.05.

That is, $P-value(=0.0574)> Significance level(=0.05)$ .

Therefore, it can be concluded that the null hypothesis is not rejected at the 0.05 significance level.

Hence, there is no evidence that the proportion of time the coin landed as heads up is different from 0.5.

In this context, even if the significance is increased, there is no change in the conclusion.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

Check whether the sample data provide convincing evidence that the proportion of time a coin landed as heads up is different from 0.5 at 0.01 significance level.

Answer to Problem 88CR

No, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

Explanation of Solution

Calculation:

In a sample of 250 spinning of Belgium-minted Euro coin, 140 spins were founded to turn up as heads.

Step 1:

Population characteristic of interest:

Let p be the population characteristic of interest.

In this context, p denotes the population proportion of spins that landed as heads up.

Step 2:

Null hypothesis: $H0:p=0.5$

That is, the proportion of spins that landed as heads up is 0.5.

Step 3:

Alternative hypothesis: $Ha:p≠0.5$

That is, the proportion of spins that landed as heads up is different from 0.5.

Step 4:

Significance level, $α$ :

It is given that the significance level, $α=0.01$ .

Step 5:

Test statistic, z:

$z=p^−pp(1−p)n$ ,

Where, $p^$ be the sample proportion, p is the hypothesized proportion, and n is the sample size.

Substitute the hypothesized proportion, $p=0.5$ , in the test statistic.

$z=p^−0.50.5(1−0.5)n$

Here, the sample proportion $p^$ is not known.

Step 6:

Assumptions:

Let $p^$ be the sample proportion from a random sample.
The large sample z test can be used if the sample size n satisfies the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .
The sample size should not be greater than 10% of the population size.

Requirement check:

It is assumed that the sample is from a random sample.
Check the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .

$n(hypothesized value)=np=250(0.5)=175>10$

$n(1−hypothesized value)=n(1−p)=250(1−0.5)=250(0.5)=175>10$

Since $n(hypothesized value)$ and $n(1−hypothesized value)$ are greater than 10, the sample size requirement is met.

Therefore, the large sample z test is appropriate.

Although, the population size is not known, it is reasonable to assume that the sample size of 250 acts as the representative for all the spins of Euro coins. It is also definite that the sample size is less than the 10% of the population of the spins of Euro coins.

Step 7:

The value of the test statistic is obtained as follows:

$z=p^−0.50.5(1−0.5)n$

The sample proportion, $p^$ , is obtained as follows:

$p^=xn$

Where, x is the number of observations of interest and n is the sample size.

Substitute the corresponding values to get the sample proportion.

$p^=140250=0.56$

By substituting the value of sample proportion, the test statistic is obtained as follows:

$z=0.56−0.50.5(1−0.5)250=0.060.032≈1.90$

Thus, the value of test statistic is 1.90.

Step 8:

P-value:

In this context, the alternative hypothesis denotes that the test carried out is two-tailed test. Therefore, the P-value is the area under the z curve and the two-tailed of the calculated z value.

The P-value for the test statistic value of 1.90 is obtained as follows:

$P-value=2×{1−Area to the Right of 1.90}=2×{1−P(z<1.90)}$

Use standard normal probabilities (Cumulative z curve areas) table to find the z-value.

Procedure:

For z at 1.90:

 Locate 1.9 in the left column of the table.

 Obtain the value in the corresponding row below .00.

That is, $P(z<1.90)=0.9713$ .

The probability of the event is as given below:

$P-value=2{1−P(z<1.90)}=2{1−0.9713}=2{0.0287}=0.0574$

Thus, the P-value for the test statistic of 1.90 is 0.0574.

Step 9:

Decision rule:

If $P-value≤Significance level$ , then reject the null hypothesis $H0$ .

If $P-value>Significance level$ , then fail to reject the null hypothesis $H0$ .

Here, the $P-value$ of 0.0574 is greater than the significance level 0.01.

That is, $P-value(=0.0574)>Significance level(=0.01)$ .

The decision is that the null hypothesis is not rejected.

Conclusion:

Hence, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

b.

Expert Solution

To determine

Explain whether the conclusion for the hypothesis test is different at significance level $α=0.05$ .

Explanation of Solution

Increasing significance level increases, the chance to reject the null hypothesis.

It is given that the significance level $α=0.05$ .

Here, the $P-value$ of 0.0574 is less than the significance level 0.05.

That is, $P-value(=0.0574)> Significance level(=0.05)$ .

Therefore, it can be concluded that the null hypothesis is not rejected at the 0.05 significance level.

Hence, there is no evidence that the proportion of time the coin landed as heads up is different from 0.5.

In this context, even if the significance is increased, there is no change in the conclusion.

Answer 8

a.

Expert Solution

To determine

Check whether the sample data provide convincing evidence that the proportion of time a coin landed as heads up is different from 0.5 at 0.01 significance level.

Answer to Problem 88CR

No, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

Explanation of Solution

Calculation:

In a sample of 250 spinning of Belgium-minted Euro coin, 140 spins were founded to turn up as heads.

Step 1:

Population characteristic of interest:

Let p be the population characteristic of interest.

In this context, p denotes the population proportion of spins that landed as heads up.

Step 2:

Null hypothesis: $H0:p=0.5$

That is, the proportion of spins that landed as heads up is 0.5.

Step 3:

Alternative hypothesis: $Ha:p≠0.5$

That is, the proportion of spins that landed as heads up is different from 0.5.

Step 4:

Significance level, $α$ :

It is given that the significance level, $α=0.01$ .

Step 5:

Test statistic, z:

$z=p^−pp(1−p)n$ ,

Where, $p^$ be the sample proportion, p is the hypothesized proportion, and n is the sample size.

Substitute the hypothesized proportion, $p=0.5$ , in the test statistic.

$z=p^−0.50.5(1−0.5)n$

Here, the sample proportion $p^$ is not known.

Step 6:

Assumptions:

Let $p^$ be the sample proportion from a random sample.
The large sample z test can be used if the sample size n satisfies the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .
The sample size should not be greater than 10% of the population size.

Requirement check:

It is assumed that the sample is from a random sample.
Check the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .

$n(hypothesized value)=np=250(0.5)=175>10$

$n(1−hypothesized value)=n(1−p)=250(1−0.5)=250(0.5)=175>10$

Since $n(hypothesized value)$ and $n(1−hypothesized value)$ are greater than 10, the sample size requirement is met.

Therefore, the large sample z test is appropriate.

Although, the population size is not known, it is reasonable to assume that the sample size of 250 acts as the representative for all the spins of Euro coins. It is also definite that the sample size is less than the 10% of the population of the spins of Euro coins.

Step 7:

The value of the test statistic is obtained as follows:

$z=p^−0.50.5(1−0.5)n$

The sample proportion, $p^$ , is obtained as follows:

$p^=xn$

Where, x is the number of observations of interest and n is the sample size.

Substitute the corresponding values to get the sample proportion.

$p^=140250=0.56$

By substituting the value of sample proportion, the test statistic is obtained as follows:

$z=0.56−0.50.5(1−0.5)250=0.060.032≈1.90$

Thus, the value of test statistic is 1.90.

Step 8:

P-value:

In this context, the alternative hypothesis denotes that the test carried out is two-tailed test. Therefore, the P-value is the area under the z curve and the two-tailed of the calculated z value.

The P-value for the test statistic value of 1.90 is obtained as follows:

$P-value=2×{1−Area to the Right of 1.90}=2×{1−P(z<1.90)}$

Use standard normal probabilities (Cumulative z curve areas) table to find the z-value.

Procedure:

For z at 1.90:

 Locate 1.9 in the left column of the table.

 Obtain the value in the corresponding row below .00.

That is, $P(z<1.90)=0.9713$ .

The probability of the event is as given below:

$P-value=2{1−P(z<1.90)}=2{1−0.9713}=2{0.0287}=0.0574$

Thus, the P-value for the test statistic of 1.90 is 0.0574.

Step 9:

Decision rule:

If $P-value≤Significance level$ , then reject the null hypothesis $H0$ .

If $P-value>Significance level$ , then fail to reject the null hypothesis $H0$ .

Here, the $P-value$ of 0.0574 is greater than the significance level 0.01.

That is, $P-value(=0.0574)>Significance level(=0.01)$ .

The decision is that the null hypothesis is not rejected.

Conclusion:

Hence, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Check whether the sample data provide convincing evidence that the proportion of time a coin landed as heads up is different from 0.5 at 0.01 significance level.

Answer 13

Answer to Problem 88CR

No, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

Answer 14

Explanation of Solution

Calculation:

In a sample of 250 spinning of Belgium-minted Euro coin, 140 spins were founded to turn up as heads.

Step 1:

Population characteristic of interest:

Let p be the population characteristic of interest.

In this context, p denotes the population proportion of spins that landed as heads up.

Step 2:

Null hypothesis: $H0:p=0.5$

That is, the proportion of spins that landed as heads up is 0.5.

Step 3:

Alternative hypothesis: $Ha:p≠0.5$

That is, the proportion of spins that landed as heads up is different from 0.5.

Step 4:

Significance level, $α$ :

It is given that the significance level, $α=0.01$ .

Step 5:

Test statistic, z:

$z=p^−pp(1−p)n$ ,

Where, $p^$ be the sample proportion, p is the hypothesized proportion, and n is the sample size.

Substitute the hypothesized proportion, $p=0.5$ , in the test statistic.

$z=p^−0.50.5(1−0.5)n$

Here, the sample proportion $p^$ is not known.

Step 6:

Assumptions:

Let $p^$ be the sample proportion from a random sample.
The large sample z test can be used if the sample size n satisfies the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .
The sample size should not be greater than 10% of the population size.

Requirement check:

It is assumed that the sample is from a random sample.
Check the conditions: $n(hypothesized value)≥10$ and $n(1−hypothesized value)≥10$ .

$n(hypothesized value)=np=250(0.5)=175>10$

$n(1−hypothesized value)=n(1−p)=250(1−0.5)=250(0.5)=175>10$

Since $n(hypothesized value)$ and $n(1−hypothesized value)$ are greater than 10, the sample size requirement is met.

Therefore, the large sample z test is appropriate.

Although, the population size is not known, it is reasonable to assume that the sample size of 250 acts as the representative for all the spins of Euro coins. It is also definite that the sample size is less than the 10% of the population of the spins of Euro coins.

Step 7:

The value of the test statistic is obtained as follows:

$z=p^−0.50.5(1−0.5)n$

The sample proportion, $p^$ , is obtained as follows:

$p^=xn$

Where, x is the number of observations of interest and n is the sample size.

Substitute the corresponding values to get the sample proportion.

$p^=140250=0.56$

By substituting the value of sample proportion, the test statistic is obtained as follows:

$z=0.56−0.50.5(1−0.5)250=0.060.032≈1.90$

Thus, the value of test statistic is 1.90.

Step 8:

P-value:

In this context, the alternative hypothesis denotes that the test carried out is two-tailed test. Therefore, the P-value is the area under the z curve and the two-tailed of the calculated z value.

The P-value for the test statistic value of 1.90 is obtained as follows:

$P-value=2×{1−Area to the Right of 1.90}=2×{1−P(z<1.90)}$

Use standard normal probabilities (Cumulative z curve areas) table to find the z-value.

Procedure:

For z at 1.90:

 Locate 1.9 in the left column of the table.

 Obtain the value in the corresponding row below .00.

That is, $P(z<1.90)=0.9713$ .

The probability of the event is as given below:

$P-value=2{1−P(z<1.90)}=2{1−0.9713}=2{0.0287}=0.0574$

Thus, the P-value for the test statistic of 1.90 is 0.0574.

Step 9:

Decision rule:

If $P-value≤Significance level$ , then reject the null hypothesis $H0$ .

If $P-value>Significance level$ , then fail to reject the null hypothesis $H0$ .

Here, the $P-value$ of 0.0574 is greater than the significance level 0.01.

That is, $P-value(=0.0574)>Significance level(=0.01)$ .

The decision is that the null hypothesis is not rejected.

Conclusion:

Hence, the sample data do not provide the evidence that the proportion of time the coin landed as heads up is different from 0.5.

Answer 15

b.

Expert Solution

To determine

Explain whether the conclusion for the hypothesis test is different at significance level $α=0.05$ .

Explanation of Solution

Increasing significance level increases, the chance to reject the null hypothesis.

It is given that the significance level $α=0.05$ .

Here, the $P-value$ of 0.0574 is less than the significance level 0.05.

That is, $P-value(=0.0574)> Significance level(=0.05)$ .

Therefore, it can be concluded that the null hypothesis is not rejected at the 0.05 significance level.

Hence, there is no evidence that the proportion of time the coin landed as heads up is different from 0.5.

In this context, even if the significance is increased, there is no change in the conclusion.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Explain whether the conclusion for the hypothesis test is different at significance level $α=0.05$ .

Answer 20

Explanation of Solution

Increasing significance level increases, the chance to reject the null hypothesis.

It is given that the significance level $α=0.05$ .

Here, the $P-value$ of 0.0574 is less than the significance level 0.05.

That is, $P-value(=0.0574)> Significance level(=0.05)$ .

Therefore, it can be concluded that the null hypothesis is not rejected at the 0.05 significance level.

Hence, there is no evidence that the proportion of time the coin landed as heads up is different from 0.5.

In this context, even if the significance is increased, there is no change in the conclusion.

Answer 21

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