Math / Data & Statistics

QuestionQuestion 25 Points: 1
A house cleaning service claims that it can clean a four-bedroom house in less than 2 hours. A sample of $n=16$ houses is taken, and the sample mean is found to be 1.97 hours and the sample standard deviation is found to be 0.1 hours.
Write the interpretation base on the decision. A. There is enough evidence to reject the claim B. There is not enough evidence to support the claim C. There is enough evidence to support the claim D. There is not enough evidence to reject the claim

Studdy Solution

STEP 1

1. The null hypothesis ( $H_0$ ) is that the mean cleaning time is 2 hours or more.
2. The alternative hypothesis ( $H_a$ ) is that the mean cleaning time is less than 2 hours.
3. A significance level ( $\alpha$ ) is typically set at 0.05 unless otherwise specified.
4. The sample data follows a normal distribution or the sample size is large enough for the Central Limit Theorem to apply.

STEP 2

1. Formulate the hypotheses.
2. Calculate the test statistic.
3. Determine the critical value and decision rule.
4. Make a decision based on the test statistic and critical value.

STEP 3

Formulate the hypotheses:
- Null hypothesis ( $H_0$ ): $\mu \geq 2$ - Alternative hypothesis ( $H_a$ ): $\mu < 2$

STEP 4

Calculate the test statistic using the formula for a one-sample t-test:
$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$
where $\bar{x} = 1.97$ , $\mu_0 = 2$ , $s = 0.1$ , and $n = 16$ .
$t = \frac{1.97 - 2}{0.1/\sqrt{16}} = \frac{-0.03}{0.025} = -1.2$

STEP 5

Determine the critical value for a one-tailed t-test with $n-1 = 15$ degrees of freedom at $\alpha = 0.05$ .
Using a t-distribution table, the critical value is approximately $-1.753$ .

STEP 6

Make a decision:
- Compare the test statistic to the critical value: $-1.2$ is greater than $-1.753$ . - Since the test statistic is not in the critical region, we fail to reject the null hypothesis.
Interpretation: D. There is not enough evidence to reject the claim.

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Studdy Solution

STEP 1

STEP 2

1. Formulate the hypotheses.
2. Calculate the test statistic.
3. Determine the critical value and decision rule.
4. Make a decision based on the test statistic and critical value.

STEP 3

Formulate the hypotheses:
- Null hypothesis ( $H_0$ ): $\mu \geq 2$ - Alternative hypothesis ( $H_a$ ): $\mu < 2$

STEP 4

Calculate the test statistic using the formula for a one-sample t-test:
$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$
where $\bar{x} = 1.97$ , $\mu_0 = 2$ , $s = 0.1$ , and $n = 16$ .
$t = \frac{1.97 - 2}{0.1/\sqrt{16}} = \frac{-0.03}{0.025} = -1.2$

STEP 5

Determine the critical value for a one-tailed t-test with $n-1 = 15$ degrees of freedom at $\alpha = 0.05$ .
Using a t-distribution table, the critical value is approximately $-1.753$ .

STEP 6

Make a decision:
- Compare the test statistic to the critical value: $-1.2$ is greater than $-1.753$ . - Since the test statistic is not in the critical region, we fail to reject the null hypothesis.
Interpretation: D. There is not enough evidence to reject the claim.

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Studdy Solution

STEP 1

STEP 2

1. Formulate the hypotheses.2. Calculate the test statistic.3. Determine the critical value and decision rule.4. Make a decision based on the test statistic and critical value.

STEP 3

Formulate the hypotheses:- Null hypothesis (H0H_0H0​): μ≥2\mu \geq 2μ≥2 - Alternative hypothesis (HaH_aHa​): μ<2\mu < 2μ<2

STEP 4

STEP 5

Determine the critical value for a one-tailed t-test with n−1=15n-1 = 15n−1=15 degrees of freedom at α=0.05\alpha = 0.05α=0.05.Using a t-distribution table, the critical value is approximately −1.753-1.753−1.753.

STEP 6

Make a decision:- Compare the test statistic to the critical value: −1.2-1.2−1.2 is greater than −1.753-1.753−1.753. - Since the test statistic is not in the critical region, we fail to reject the null hypothesis.Interpretation: D. There is not enough evidence to reject the claim.

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1. Formulate the hypotheses.
2. Calculate the test statistic.
3. Determine the critical value and decision rule.
4. Make a decision based on the test statistic and critical value.

Formulate the hypotheses:
- Null hypothesis ( $H_0$ ): $\mu \geq 2$ - Alternative hypothesis ( $H_a$ ): $\mu < 2$

Determine the critical value for a one-tailed t-test with $n-1 = 15$ degrees of freedom at $\alpha = 0.05$ .
Using a t-distribution table, the critical value is approximately $-1.753$ .

Make a decision:
- Compare the test statistic to the critical value: $-1.2$ is greater than $-1.753$ . - Since the test statistic is not in the critical region, we fail to reject the null hypothesis.
Interpretation: D. There is not enough evidence to reject the claim.