Math / Data & Statistics

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A random sample of 86 eighth grade students' scores on a national mathematics assessment test has a mean score of 284. This test result prompts a state school ad to declare that the mean score for the state's eighth graders on this exam is more than 280 . Assume that the population standard deviation is 35 . At $\alpha=0.11$ , is there evidence to support the administrator's claim? Complete parts (a) through (e). (b) Find the standardized test statistic $z$ . $z=1.06$ (Round to two decimal places as needed.) (c) Find the P -value. $P$ -value $=0.145$ (Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. Reject $\mathrm{H}_{0}$ Fail to reject $\mathrm{H}_{0}$ (e) Interpret your decision in the context of the original claim.
At the $11 \%$ significance level, there 280. $\square$ enough evidence to $\square$ the administrator's claim that the mean score for the state's eighth graders on the exam is more

Studdy Solution

STEP 1

1. The sample size $n = 86$ .
2. The sample mean $\bar{x} = 284$ .
3. The population standard deviation $\sigma = 35$ .
4. The null hypothesis $H_0: \mu = 280$ .
5. The alternative hypothesis $H_a: \mu > 280$ .
6. Significance level $\alpha = 0.11$ .

STEP 2

1. Calculate the standardized test statistic $z$ .
2. Find the P-value.
3. Decide whether to reject or fail to reject the null hypothesis.
4. Interpret the decision in the context of the original claim.

STEP 3

Calculate the standardized test statistic $z$ using the formula:
$z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$
where $\bar{x} = 284$ , $\mu_0 = 280$ , $\sigma = 35$ , and $n = 86$ .
$z = \frac{284 - 280}{35 / \sqrt{86}}$
$z = \frac{4}{35 / \sqrt{86}}$
$z = 1.06$

STEP 4

Find the P-value for $z = 1.06$ .
The P-value is the probability that $Z$ is greater than 1.06 in a standard normal distribution.
$P(Z > 1.06) = 0.145$

STEP 5

Compare the P-value with the significance level $\alpha = 0.11$ .
Since $P = 0.145 > \alpha = 0.11$ , we fail to reject the null hypothesis.

STEP 6

Interpret the decision in the context of the original claim.
At the 11% significance level, there is not enough evidence to support the administrator's claim that the mean score for the state's eighth graders on the exam is more than 280.

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Studdy Solution

STEP 1

1. The sample size $n = 86$ .
2. The sample mean $\bar{x} = 284$ .
3. The population standard deviation $\sigma = 35$ .
4. The null hypothesis $H_0: \mu = 280$ .
5. The alternative hypothesis $H_a: \mu > 280$ .
6. Significance level $\alpha = 0.11$ .

STEP 2

1. Calculate the standardized test statistic $z$ .
2. Find the P-value.
3. Decide whether to reject or fail to reject the null hypothesis.
4. Interpret the decision in the context of the original claim.

STEP 3

Calculate the standardized test statistic $z$ using the formula:
$z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$
where $\bar{x} = 284$ , $\mu_0 = 280$ , $\sigma = 35$ , and $n = 86$ .
$z = \frac{284 - 280}{35 / \sqrt{86}}$
$z = \frac{4}{35 / \sqrt{86}}$
$z = 1.06$

STEP 4

Find the P-value for $z = 1.06$ .
The P-value is the probability that $Z$ is greater than 1.06 in a standard normal distribution.
$P(Z > 1.06) = 0.145$

STEP 5

Compare the P-value with the significance level $\alpha = 0.11$ .
Since $P = 0.145 > \alpha = 0.11$ , we fail to reject the null hypothesis.

STEP 6

Interpret the decision in the context of the original claim.
At the 11% significance level, there is not enough evidence to support the administrator's claim that the mean score for the state's eighth graders on the exam is more than 280.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the standardized test statistic z z z.2. Find the P-value.3. Decide whether to reject or fail to reject the null hypothesis.4. Interpret the decision in the context of the original claim.

STEP 3

STEP 4

Find the P-value for z=1.06 z = 1.06 z=1.06.The P-value is the probability that Z Z Z is greater than 1.06 in a standard normal distribution.P(Z>1.06)=0.145 P(Z > 1.06) = 0.145 P(Z>1.06)=0.145

STEP 5

Compare the P-value with the significance level α=0.11 \alpha = 0.11 α=0.11.Since P=0.145>α=0.11 P = 0.145 > \alpha = 0.11 P=0.145>α=0.11, we fail to reject the null hypothesis.

STEP 6

Interpret the decision in the context of the original claim.At the 11% significance level, there is not enough evidence to support the administrator's claim that the mean score for the state's eighth graders on the exam is more than 280.

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

1. Calculate the standardized test statistic $z$ .
2. Find the P-value.
3. Decide whether to reject or fail to reject the null hypothesis.
4. Interpret the decision in the context of the original claim.

Find the P-value for $z = 1.06$ .
The P-value is the probability that $Z$ is greater than 1.06 in a standard normal distribution.
$P(Z > 1.06) = 0.145$

Compare the P-value with the significance level $\alpha = 0.11$ .
Since $P = 0.145 > \alpha = 0.11$ , we fail to reject the null hypothesis.

Interpret the decision in the context of the original claim.
At the 11% significance level, there is not enough evidence to support the administrator's claim that the mean score for the state's eighth graders on the exam is more than 280.