Solved on Mar 03, 2024

Find the P-value for a right-tailed test with test statistic $z=0.52$ . Use a 0.05 significance level to determine if the null hypothesis should be rejected or failed to be rejected.

STEP 1

Assumptions
1. The test is a right-tailed test.
2. The test statistic given is $z = 0.52$ .
3. The significance level ( $\alpha$ ) is 0.05.

STEP 2

The P-value is the probability that the test statistic will be greater than or equal to the observed value, assuming the null hypothesis is true.

STEP 3

To find the P-value, we need to look up the test statistic $z = 0.52$ in the standard normal distribution (Z-table) or use a calculator that provides the cumulative distribution function (CDF) for the standard normal distribution.

STEP 4

The P-value is given by $P(Z \geq 0.52)$ , which is equal to $1 - P(Z < 0.52)$ .

STEP 5

Using a Z-table or calculator, find the cumulative probability for $Z < 0.52$ .

STEP 6

Subtract the cumulative probability from 1 to find the P-value.

STEP 7

If the P-value is less than the significance level ( $\alpha = 0.05$ ), we reject the null hypothesis. If the P-value is greater than or equal to the significance level, we fail to reject the null hypothesis.

STEP 8

Let's find the cumulative probability for $Z < 0.52$ using a Z-table or calculator.

STEP 9

The cumulative probability for $Z < 0.52$ is approximately 0.6985.

STEP 10

Calculate the P-value using the cumulative probability.
$P-value = 1 - 0.6985$

STEP 11

Perform the subtraction to find the P-value.
$P-value = 1 - 0.6985 = 0.3015$

STEP 12

Compare the P-value to the significance level.
$P-value = 0.3015$ $\alpha = 0.05$

STEP 13

Since $P-value > \alpha$ , we fail to reject the null hypothesis.
The P-value is 0.3015, and since it is greater than the significance level of 0.05, we fail to reject the null hypothesis.

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Find the P-value for a right-tailed test with test statistic $z=0.52$ . Use a 0.05 significance level to determine if the null hypothesis should be rejected or failed to be rejected.

STEP 1

Assumptions
1. The test is a right-tailed test.
2. The test statistic given is $z = 0.52$ .
3. The significance level ( $\alpha$ ) is 0.05.

STEP 2

The P-value is the probability that the test statistic will be greater than or equal to the observed value, assuming the null hypothesis is true.

STEP 3

To find the P-value, we need to look up the test statistic $z = 0.52$ in the standard normal distribution (Z-table) or use a calculator that provides the cumulative distribution function (CDF) for the standard normal distribution.

STEP 4

The P-value is given by $P(Z \geq 0.52)$ , which is equal to $1 - P(Z < 0.52)$ .

STEP 5

Using a Z-table or calculator, find the cumulative probability for $Z < 0.52$ .

STEP 6

Subtract the cumulative probability from 1 to find the P-value.

STEP 7

If the P-value is less than the significance level ( $\alpha = 0.05$ ), we reject the null hypothesis. If the P-value is greater than or equal to the significance level, we fail to reject the null hypothesis.

STEP 8

Let's find the cumulative probability for $Z < 0.52$ using a Z-table or calculator.

STEP 9

The cumulative probability for $Z < 0.52$ is approximately 0.6985.

STEP 10

Calculate the P-value using the cumulative probability.
$P-value = 1 - 0.6985$

STEP 11

Perform the subtraction to find the P-value.
$P-value = 1 - 0.6985 = 0.3015$

STEP 12

Compare the P-value to the significance level.
$P-value = 0.3015$ $\alpha = 0.05$

STEP 13

Since $P-value > \alpha$ , we fail to reject the null hypothesis.
The P-value is 0.3015, and since it is greater than the significance level of 0.05, we fail to reject the null hypothesis.

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Find the P-value for a right-tailed test with test statistic z=0.52z=0.52z=0.52. Use a 0.05 significance level to determine if the null hypothesis should be rejected or failed to be rejected.

STEP 1

Assumptions1. The test is a right-tailed test.2. The test statistic given is z=0.52 z = 0.52 z=0.52.3. The significance level (α \alpha α) is 0.05.

STEP 2

The P-value is the probability that the test statistic will be greater than or equal to the observed value, assuming the null hypothesis is true.

STEP 3

To find the P-value, we need to look up the test statistic z=0.52 z = 0.52 z=0.52 in the standard normal distribution (Z-table) or use a calculator that provides the cumulative distribution function (CDF) for the standard normal distribution.

STEP 4

The P-value is given by P(Z≥0.52) P(Z \geq 0.52) P(Z≥0.52), which is equal to 1−P(Z<0.52) 1 - P(Z < 0.52) 1−P(Z<0.52).

STEP 5

Using a Z-table or calculator, find the cumulative probability for Z<0.52 Z < 0.52 Z<0.52.

STEP 6

Subtract the cumulative probability from 1 to find the P-value.

STEP 7

If the P-value is less than the significance level (α=0.05 \alpha = 0.05 α=0.05), we reject the null hypothesis. If the P-value is greater than or equal to the significance level, we fail to reject the null hypothesis.

STEP 8

Let's find the cumulative probability for Z<0.52 Z < 0.52 Z<0.52 using a Z-table or calculator.

STEP 9

The cumulative probability for Z<0.52 Z < 0.52 Z<0.52 is approximately 0.6985.

STEP 10

Calculate the P-value using the cumulative probability.P−value=1−0.6985 P-value = 1 - 0.6985 P−value=1−0.6985

STEP 11

Perform the subtraction to find the P-value.P−value=1−0.6985=0.3015 P-value = 1 - 0.6985 = 0.3015 P−value=1−0.6985=0.3015

STEP 12

Compare the P-value to the significance level.P−value=0.3015 P-value = 0.3015 P−value=0.3015 α=0.05 \alpha = 0.05 α=0.05

STEP 13

Since P−value>α P-value > \alpha P−value>α, we fail to reject the null hypothesis.The P-value is 0.3015, and since it is greater than the significance level of 0.05, we fail to reject the null hypothesis.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Find the P-value for a right-tailed test with test statistic $z=0.52$ . Use a 0.05 significance level to determine if the null hypothesis should be rejected or failed to be rejected.

Assumptions
1. The test is a right-tailed test.
2. The test statistic given is $z = 0.52$ .
3. The significance level ( $\alpha$ ) is 0.05.

To find the P-value, we need to look up the test statistic $z = 0.52$ in the standard normal distribution (Z-table) or use a calculator that provides the cumulative distribution function (CDF) for the standard normal distribution.

The P-value is given by $P(Z \geq 0.52)$ , which is equal to $1 - P(Z < 0.52)$ .

Using a Z-table or calculator, find the cumulative probability for $Z < 0.52$ .

If the P-value is less than the significance level ( $\alpha = 0.05$ ), we reject the null hypothesis. If the P-value is greater than or equal to the significance level, we fail to reject the null hypothesis.

Let's find the cumulative probability for $Z < 0.52$ using a Z-table or calculator.

The cumulative probability for $Z < 0.52$ is approximately 0.6985.

Calculate the P-value using the cumulative probability.
$P-value = 1 - 0.6985$

Perform the subtraction to find the P-value.
$P-value = 1 - 0.6985 = 0.3015$

Compare the P-value to the significance level.
$P-value = 0.3015$ $\alpha = 0.05$

Since $P-value > \alpha$ , we fail to reject the null hypothesis.
The P-value is 0.3015, and since it is greater than the significance level of 0.05, we fail to reject the null hypothesis.