Math / Data & Statistics

QuestionListed below are the lead concentrations (in $\mu \mathrm{g} / \mathrm{g}$ ) measured in different Ayurveda medicines. Ayurveda is a traditional medical system commonly used in India. The lead concentrations listed here are from medicines manufactured in the United States. Assume that a simple random sample has been selected. Use a 0.10 significance level to test the claim that the mean lead concentration for all such medicines is less than $14.0 \mu \mathrm{~g} / \mathrm{g}$ . $\begin{array}{lllllllllll} 3.01 & 6.45 & 6.04 & 5.49 & 20.50 & 7.46 & 12.02 & 20.51 & 11.49 & 17.52 & \square \end{array}$
Identify the null and alternative hypotheses. $\begin{array}{l} H_{0}: \mu=14.0 \\ H_{1}: \mu=14.0 \end{array}$ (Type integers or decimals. Do not round.) Identify the test statistic. $-1.44$ (Round to two decimal places as needed.) Identify the P -value. $\square$ (Round to three decimal places as needed.)

Studdy Solution

STEP 1

1. The sample is a simple random sample of lead concentrations in Ayurveda medicines.
2. The sample size is $n = 10$ .
3. The population standard deviation is unknown, so a t-test is appropriate.
4. The significance level is $\alpha = 0.10$ .

STEP 2

1. Identify the null and alternative hypotheses.
2. Calculate the test statistic.
3. Determine the P-value.
4. Make a conclusion based on the P-value and significance level.

STEP 3

Identify the null and alternative hypotheses.
$H_0: \mu = 14.0$
$H_1: \mu < 14.0$

STEP 4

The test statistic is given as $-1.44$ .

STEP 5

Determine the P-value using the t-distribution with $n-1 = 9$ degrees of freedom.
Using a t-table or calculator, find the P-value for $t = -1.44$ with 9 degrees of freedom.
The P-value is approximately $0.089$ .

STEP 6

Compare the P-value to the significance level $\alpha = 0.10$ .
Since the P-value $0.089$ is less than $0.10$ , we reject the null hypothesis.
Conclusion: There is sufficient evidence to support the claim that the mean lead concentration is less than $14.0 \mu \mathrm{g} / \mathrm{g}$ .
The P-value is:
$\boxed{0.089}$

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Studdy Solution

STEP 1

1. The sample is a simple random sample of lead concentrations in Ayurveda medicines.
2. The sample size is $n = 10$ .
3. The population standard deviation is unknown, so a t-test is appropriate.
4. The significance level is $\alpha = 0.10$ .

STEP 2

1. Identify the null and alternative hypotheses.
2. Calculate the test statistic.
3. Determine the P-value.
4. Make a conclusion based on the P-value and significance level.

STEP 3

Identify the null and alternative hypotheses.
$H_0: \mu = 14.0$
$H_1: \mu < 14.0$

STEP 4

The test statistic is given as $-1.44$ .

STEP 5

Determine the P-value using the t-distribution with $n-1 = 9$ degrees of freedom.
Using a t-table or calculator, find the P-value for $t = -1.44$ with 9 degrees of freedom.
The P-value is approximately $0.089$ .

STEP 6

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Studdy Solution

STEP 1

1. The sample is a simple random sample of lead concentrations in Ayurveda medicines.2. The sample size is n=10 n = 10 n=10.3. The population standard deviation is unknown, so a t-test is appropriate.4. The significance level is α=0.10 \alpha = 0.10 α=0.10.

STEP 2

1. Identify the null and alternative hypotheses.2. Calculate the test statistic.3. Determine the P-value.4. Make a conclusion based on the P-value and significance level.

STEP 3

Identify the null and alternative hypotheses.H0:μ=14.0H_0: \mu = 14.0H0​:μ=14.0H1:μ<14.0H_1: \mu < 14.0H1​:μ<14.0

STEP 4

The test statistic is given as −1.44-1.44−1.44.

STEP 5

Determine the P-value using the t-distribution with n−1=9 n-1 = 9 n−1=9 degrees of freedom.Using a t-table or calculator, find the P-value for t=−1.44 t = -1.44 t=−1.44 with 9 degrees of freedom.The P-value is approximately 0.089 0.089 0.089.

STEP 6

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1. The sample is a simple random sample of lead concentrations in Ayurveda medicines.
2. The sample size is $n = 10$ .
3. The population standard deviation is unknown, so a t-test is appropriate.
4. The significance level is $\alpha = 0.10$ .

1. Identify the null and alternative hypotheses.
2. Calculate the test statistic.
3. Determine the P-value.
4. Make a conclusion based on the P-value and significance level.

Identify the null and alternative hypotheses.
$H_0: \mu = 14.0$
$H_1: \mu < 14.0$

The test statistic is given as $-1.44$ .

Determine the P-value using the t-distribution with $n-1 = 9$ degrees of freedom.
Using a t-table or calculator, find the P-value for $t = -1.44$ with 9 degrees of freedom.
The P-value is approximately $0.089$ .