Math / Data & Statistics

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A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. Use a 0.01 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one minute? $\begin{array}{lllllllllllllll} 76 & 93 & 47 & 76 & 48 & 35 & 72 & 70 & 77 & 59 & 74 & 80 & 102 & 95 & 71 \end{array}$
Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? A. $\mathrm{H}_{0}: \mu=60$ seconds B. $H_{0}: \mu=60$ seconds $H_{1}: \mu>60$ seconds $H_{1}: \mu \neq 60$ seconds C. $\mathrm{H}_{0}: \mu=60$ seconds D. $H_{0}: \mu \neq 60$ seconds $\mathrm{H}_{1}: \mu<60$ seconds $H_{1}: \mu=60$ seconds
Determine the test statistic. (Round to two decimal places as needed.) Determine the P -value. (Round to three decimal places as needed.) State the final conclusion that addresses the original claim.

Studdy Solution

STEP 1

1. The sample data consists of 15 times in seconds: $76, 93, 47, 76, 48, 35, 72, 70, 77, 59, 74, 80, 102, 95, 71$ .
2. We are testing the claim that the population mean is 60 seconds.
3. A significance level of 0.01 is used.
4. The sample is assumed to be drawn from a normally distributed population or the sample size is sufficiently large for the Central Limit Theorem to apply.

STEP 2

1. Formulate the null and alternative hypotheses.
2. Calculate the sample mean and standard deviation.
3. Determine the test statistic.
4. Calculate the P-value.
5. State the conclusion based on the P-value and significance level.

STEP 3

Formulate the hypotheses:
- Null hypothesis ( $H_0$ ): The population mean is equal to 60 seconds. - Alternative hypothesis ( $H_1$ ): The population mean is not equal to 60 seconds.
$H_0: \mu = 60$ $H_1: \mu \neq 60$

STEP 4

Calculate the sample mean ( $\bar{x}$ ) and standard deviation ( $s$ ).
Given data: $76, 93, 47, 76, 48, 35, 72, 70, 77, 59, 74, 80, 102, 95, 71$ .
Calculate $\bar{x}$ :
$\bar{x} = \frac{76 + 93 + 47 + 76 + 48 + 35 + 72 + 70 + 77 + 59 + 74 + 80 + 102 + 95 + 71}{15} = \frac{1175}{15} = 78.33$
Calculate $s$ (standard deviation):
$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$
\[ s = \sqrt{\frac{(76-78.33)^2 + (93-78.33)^2 + \ldots + (71-78.33)^2}{14}} $
\[ s \approx 18.53 $

STEP 5

Determine the test statistic ( $t$ ) using the formula:
$t = \frac{\bar{x} - \mu}{s/\sqrt{n}}$
Where $\mu = 60$ , $n = 15$ .
$t = \frac{78.33 - 60}{18.53/\sqrt{15}} \approx \frac{18.33}{4.78} \approx 3.83$

STEP 6

Calculate the P-value for the two-tailed test using the t-distribution with $n-1 = 14$ degrees of freedom.
Using a t-table or calculator, find the P-value for $t = 3.83$ .
\[ \text{P-value} \approx 0.002 $

STEP 7

State the conclusion:
Since the P-value ( $0.002$ ) is less than the significance level ( $0.01$ ), we reject the null hypothesis.
Conclusion: There is sufficient evidence to suggest that the mean time estimated by students is not equal to 60 seconds. It appears that students are not reasonably good at estimating one minute.

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Studdy Solution

STEP 1

STEP 2

1. Formulate the null and alternative hypotheses.
2. Calculate the sample mean and standard deviation.
3. Determine the test statistic.
4. Calculate the P-value.
5. State the conclusion based on the P-value and significance level.

STEP 3

Formulate the hypotheses:
- Null hypothesis ( $H_0$ ): The population mean is equal to 60 seconds. - Alternative hypothesis ( $H_1$ ): The population mean is not equal to 60 seconds.
$H_0: \mu = 60$ $H_1: \mu \neq 60$

STEP 4

STEP 5

Determine the test statistic ( $t$ ) using the formula:
$t = \frac{\bar{x} - \mu}{s/\sqrt{n}}$
Where $\mu = 60$ , $n = 15$ .
$t = \frac{78.33 - 60}{18.53/\sqrt{15}} \approx \frac{18.33}{4.78} \approx 3.83$

STEP 6

Calculate the P-value for the two-tailed test using the t-distribution with $n-1 = 14$ degrees of freedom.
Using a t-table or calculator, find the P-value for $t = 3.83$ .
\[ \text{P-value} \approx 0.002 $

STEP 7

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Studdy Solution

STEP 1

STEP 2

1. Formulate the null and alternative hypotheses.2. Calculate the sample mean and standard deviation.3. Determine the test statistic.4. Calculate the P-value.5. State the conclusion based on the P-value and significance level.

STEP 3

Formulate the hypotheses:- Null hypothesis (H0H_0H0​): The population mean is equal to 60 seconds. - Alternative hypothesis (H1H_1H1​): The population mean is not equal to 60 seconds.H0:μ=60 H_0: \mu = 60 H0​:μ=60 H1:μ≠60 H_1: \mu \neq 60 H1​:μ=60

STEP 4

STEP 5

STEP 6

Calculate the P-value for the two-tailed test using the t-distribution with n−1=14n-1 = 14n−1=14 degrees of freedom.Using a t-table or calculator, find the P-value for t=3.83t = 3.83t=3.83.\[ \text{P-value} \approx 0.002 $

STEP 7

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

1. Formulate the null and alternative hypotheses.
2. Calculate the sample mean and standard deviation.
3. Determine the test statistic.
4. Calculate the P-value.
5. State the conclusion based on the P-value and significance level.

Formulate the hypotheses:
- Null hypothesis ( $H_0$ ): The population mean is equal to 60 seconds. - Alternative hypothesis ( $H_1$ ): The population mean is not equal to 60 seconds.
$H_0: \mu = 60$ $H_1: \mu \neq 60$

Calculate the P-value for the two-tailed test using the t-distribution with $n-1 = 14$ degrees of freedom.
Using a t-table or calculator, find the P-value for $t = 3.83$ .
\[ \text{P-value} \approx 0.002 $