Math / Data & Statistics

QuestionA company that makes cola drinks states that the mean caffeine content per 12 -ounce bottle of cola is 40 milligrams. You want to test this claim. During your tests, you find that a random sample of thirty 12-ounce bottles of cola has a mean caffeine content of 37.8 milligrams. Assume the population is normally distributed and the population standard deviation is 6.6 milligrams. At $\alpha=0.05$ , can you reject the company's claim? Complete parts (a) through (e). (d) Decide whether to reject or fail to reject the null hypothesis. A. Since $z$ is in the rejection region, fail to reject the null hypothesis. B. Since $z$ is not in the rejection region, reject the null hypothesis. C. Since $z$ is in the rejection region, reject the null hypothesis. D. Since $z$ is not in the rejection region, fail to reject the null hypothesis. (e) Interpret the decision in the context of the original claim.
At the $5 \%$ significance level, there $\square$ enough evidence to $\square$ the company's claim that the mean caffeine content per 12-ounce bottle of cola $\square$ $\square$ milligrams.

Studdy Solution

STEP 1

1. The mean caffeine content claimed by the company is 40 milligrams.
2. The sample mean caffeine content is 37.8 milligrams.
3. The population standard deviation is 6.6 milligrams.
4. The sample size is 30.
5. The significance level $\alpha$ is 0.05.
6. The population is normally distributed.

STEP 2

1. State the null and alternative hypotheses.
2. Calculate the test statistic.
3. Determine the critical value and rejection region.
4. Decide whether to reject or fail to reject the null hypothesis.
5. Interpret the decision in the context of the original claim.

STEP 3

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The mean caffeine content is 40 milligrams ( $\mu = 40$ ). - Alternative hypothesis ( $H_a$ ): The mean caffeine content is not 40 milligrams ( $\mu \neq 40$ ).

STEP 4

Calculate the test statistic using the formula:
$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$
where: - $\bar{x} = 37.8$ (sample mean), - $\mu = 40$ (population mean), - $\sigma = 6.6$ (population standard deviation), - $n = 30$ (sample size).
$z = \frac{37.8 - 40}{\frac{6.6}{\sqrt{30}}}$
$z = \frac{-2.2}{1.204} \approx -1.83$

STEP 5

Determine the critical value for a two-tailed test at $\alpha = 0.05$ :
- The critical z-values for a two-tailed test at $\alpha = 0.05$ are approximately $-1.96$ and $1.96$ .
Rejection region: $z < -1.96$ or $z > 1.96$ .

STEP 6

Decide whether to reject or fail to reject the null hypothesis:
- The calculated $z$ value is $-1.83$ , which is not in the rejection region $(-1.96, 1.96)$ .
Therefore, we fail to reject the null hypothesis.

STEP 7

Interpret the decision in the context of the original claim:
At the $5\%$ significance level, there is not enough evidence to reject the company's claim that the mean caffeine content per 12-ounce bottle of cola is 40 milligrams.

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Studdy Solution

STEP 1

1. The mean caffeine content claimed by the company is 40 milligrams.
2. The sample mean caffeine content is 37.8 milligrams.
3. The population standard deviation is 6.6 milligrams.
4. The sample size is 30.
5. The significance level $\alpha$ is 0.05.
6. The population is normally distributed.

STEP 2

1. State the null and alternative hypotheses.
2. Calculate the test statistic.
3. Determine the critical value and rejection region.
4. Decide whether to reject or fail to reject the null hypothesis.
5. Interpret the decision in the context of the original claim.

STEP 3

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The mean caffeine content is 40 milligrams ( $\mu = 40$ ). - Alternative hypothesis ( $H_a$ ): The mean caffeine content is not 40 milligrams ( $\mu \neq 40$ ).

STEP 4

STEP 5

Determine the critical value for a two-tailed test at $\alpha = 0.05$ :
- The critical z-values for a two-tailed test at $\alpha = 0.05$ are approximately $-1.96$ and $1.96$ .
Rejection region: $z < -1.96$ or $z > 1.96$ .

STEP 6

Decide whether to reject or fail to reject the null hypothesis:
- The calculated $z$ value is $-1.83$ , which is not in the rejection region $(-1.96, 1.96)$ .
Therefore, we fail to reject the null hypothesis.

STEP 7

Interpret the decision in the context of the original claim:
At the $5\%$ significance level, there is not enough evidence to reject the company's claim that the mean caffeine content per 12-ounce bottle of cola is 40 milligrams.

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Studdy Solution

STEP 1

1. The mean caffeine content claimed by the company is 40 milligrams.2. The sample mean caffeine content is 37.8 milligrams.3. The population standard deviation is 6.6 milligrams.4. The sample size is 30.5. The significance level α\alphaα is 0.05.6. The population is normally distributed.

STEP 2

1. State the null and alternative hypotheses.2. Calculate the test statistic.3. Determine the critical value and rejection region.4. Decide whether to reject or fail to reject the null hypothesis.5. Interpret the decision in the context of the original claim.

STEP 3

State the null and alternative hypotheses:- Null hypothesis (H0H_0H0​): The mean caffeine content is 40 milligrams (μ=40\mu = 40μ=40). - Alternative hypothesis (HaH_aHa​): The mean caffeine content is not 40 milligrams (μ≠40\mu \neq 40μ=40).

STEP 4

STEP 5

Determine the critical value for a two-tailed test at α=0.05\alpha = 0.05α=0.05:- The critical z-values for a two-tailed test at α=0.05\alpha = 0.05α=0.05 are approximately −1.96-1.96−1.96 and 1.961.961.96.Rejection region: z<−1.96z < -1.96z<−1.96 or z>1.96z > 1.96z>1.96.

STEP 6

Decide whether to reject or fail to reject the null hypothesis:- The calculated zzz value is −1.83-1.83−1.83, which is not in the rejection region (−1.96,1.96)(-1.96, 1.96)(−1.96,1.96).Therefore, we fail to reject the null hypothesis.

STEP 7

Interpret the decision in the context of the original claim:At the 5%5\%5% significance level, there is not enough evidence to reject the company's claim that the mean caffeine content per 12-ounce bottle of cola is 40 milligrams.

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1. The mean caffeine content claimed by the company is 40 milligrams.
2. The sample mean caffeine content is 37.8 milligrams.
3. The population standard deviation is 6.6 milligrams.
4. The sample size is 30.
5. The significance level $\alpha$ is 0.05.
6. The population is normally distributed.

1. State the null and alternative hypotheses.
2. Calculate the test statistic.
3. Determine the critical value and rejection region.
4. Decide whether to reject or fail to reject the null hypothesis.
5. Interpret the decision in the context of the original claim.

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The mean caffeine content is 40 milligrams ( $\mu = 40$ ). - Alternative hypothesis ( $H_a$ ): The mean caffeine content is not 40 milligrams ( $\mu \neq 40$ ).

Determine the critical value for a two-tailed test at $\alpha = 0.05$ :
- The critical z-values for a two-tailed test at $\alpha = 0.05$ are approximately $-1.96$ and $1.96$ .
Rejection region: $z < -1.96$ or $z > 1.96$ .

Decide whether to reject or fail to reject the null hypothesis:
- The calculated $z$ value is $-1.83$ , which is not in the rejection region $(-1.96, 1.96)$ .
Therefore, we fail to reject the null hypothesis.

Interpret the decision in the context of the original claim:
At the $5\%$ significance level, there is not enough evidence to reject the company's claim that the mean caffeine content per 12-ounce bottle of cola is 40 milligrams.