Math / Data & Statistics

QuestioniClicker Question A scientist is concerned about radiation levels in her laboratory. A room is only considered safe if the mean radiation level is 425 or less. A random sample of 16 radiation measurements is taken at different locations within the laboratory. These 16 measurements have a mean of 437 and a standard deviation of 20. Radiation levels in the laboratory are known to follow a normal distribution. We conduct a hypothesis test at the $5 \%$ level of significance to determine whether there is evidence that the laboratory is unsafe.

Studdy Solution

STEP 1

1. The sample size is $n = 16$ .
2. The sample mean is $\bar{x} = 437$ .
3. The sample standard deviation is $s = 20$ .
4. The population of radiation levels follows a normal distribution.
5. We are conducting a hypothesis test at the $5\%$ level of significance.

STEP 2

1. State the null and alternative hypotheses.
2. Determine the test statistic.
3. Determine the critical value or p-value.
4. Make a decision to accept or reject the null hypothesis.

STEP 3

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The mean radiation level is $\mu = 425$ . - Alternative hypothesis ( $H_a$ ): The mean radiation level is $\mu > 425$ .

STEP 4

Calculate the test statistic using the formula for the t-test:
$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$
Substitute the given values:
$t = \frac{437 - 425}{20/\sqrt{16}} = \frac{12}{5} = 2.4$

STEP 5

Determine the critical value or p-value for a one-tailed test at the $5\%$ level of significance with $n - 1 = 15$ degrees of freedom.
Using a t-distribution table or calculator, find the critical value for $\alpha = 0.05$ .
The critical value is approximately $t_{0.05, 15} = 1.753$ .

STEP 6

Compare the test statistic to the critical value:
Since $t = 2.4$ is greater than $1.753$ , we reject the null hypothesis.
Conclusion: There is evidence at the $5\%$ level of significance to suggest that the laboratory is unsafe.

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Studdy Solution

STEP 1

1. The sample size is $n = 16$ .
2. The sample mean is $\bar{x} = 437$ .
3. The sample standard deviation is $s = 20$ .
4. The population of radiation levels follows a normal distribution.
5. We are conducting a hypothesis test at the $5\%$ level of significance.

STEP 2

1. State the null and alternative hypotheses.
2. Determine the test statistic.
3. Determine the critical value or p-value.
4. Make a decision to accept or reject the null hypothesis.

STEP 3

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The mean radiation level is $\mu = 425$ . - Alternative hypothesis ( $H_a$ ): The mean radiation level is $\mu > 425$ .

STEP 4

Calculate the test statistic using the formula for the t-test:
$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$
Substitute the given values:
$t = \frac{437 - 425}{20/\sqrt{16}} = \frac{12}{5} = 2.4$

STEP 5

Determine the critical value or p-value for a one-tailed test at the $5\%$ level of significance with $n - 1 = 15$ degrees of freedom.
Using a t-distribution table or calculator, find the critical value for $\alpha = 0.05$ .
The critical value is approximately $t_{0.05, 15} = 1.753$ .

STEP 6

Compare the test statistic to the critical value:
Since $t = 2.4$ is greater than $1.753$ , we reject the null hypothesis.
Conclusion: There is evidence at the $5\%$ level of significance to suggest that the laboratory is unsafe.

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Studdy Solution

STEP 1

1. The sample size is n=16 n = 16 n=16.2. The sample mean is xˉ=437 \bar{x} = 437 xˉ=437.3. The sample standard deviation is s=20 s = 20 s=20.4. The population of radiation levels follows a normal distribution.5. We are conducting a hypothesis test at the 5% 5\% 5% level of significance.

STEP 2

1. State the null and alternative hypotheses.2. Determine the test statistic.3. Determine the critical value or p-value.4. Make a decision to accept or reject the null hypothesis.

STEP 3

State the null and alternative hypotheses:- Null hypothesis (H0 H_0 H0​): The mean radiation level is μ=425 \mu = 425 μ=425. - Alternative hypothesis (Ha H_a Ha​): The mean radiation level is μ>425 \mu > 425 μ>425.

STEP 4

Calculate the test statistic using the formula for the t-test:t=xˉ−μ0s/nt = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}t=s/n​xˉ−μ0​​Substitute the given values:t=437−42520/16=125=2.4t = \frac{437 - 425}{20/\sqrt{16}} = \frac{12}{5} = 2.4t=20/16​437−425​=512​=2.4

STEP 5

STEP 6

Compare the test statistic to the critical value:Since t=2.4 t = 2.4 t=2.4 is greater than 1.753 1.753 1.753, we reject the null hypothesis. Conclusion: There is evidence at the 5% 5\% 5% level of significance to suggest that the laboratory is unsafe.

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

1. The sample size is $n = 16$ .
2. The sample mean is $\bar{x} = 437$ .
3. The sample standard deviation is $s = 20$ .
4. The population of radiation levels follows a normal distribution.
5. We are conducting a hypothesis test at the $5\%$ level of significance.

1. State the null and alternative hypotheses.
2. Determine the test statistic.
3. Determine the critical value or p-value.
4. Make a decision to accept or reject the null hypothesis.

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The mean radiation level is $\mu = 425$ . - Alternative hypothesis ( $H_a$ ): The mean radiation level is $\mu > 425$ .

Calculate the test statistic using the formula for the t-test:
$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$
Substitute the given values:
$t = \frac{437 - 425}{20/\sqrt{16}} = \frac{12}{5} = 2.4$

Compare the test statistic to the critical value:
Since $t = 2.4$ is greater than $1.753$ , we reject the null hypothesis.
Conclusion: There is evidence at the $5\%$ level of significance to suggest that the laboratory is unsafe.