Questions: You wish to test the following claim ( H₁ ) at a significance level of α=0.05. H₀: μ=56.1 H₁: μ ≠ 56.1 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n=21 with mean x̄=61.8 and a standard deviation of s=6.2. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p -value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) α greater than α This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the population mean is not equal to 56.1. There is not sufficient evidence to warrant rejection of the claim that the population mean is not

You wish to test the following claim ( H₁ ) at a significance level of α=0.05. H₀: μ=56.1 H₁: μ ≠ 56.1 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n=21 with mean x̄=61.8 and a standard deviation of s=6.2. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p -value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) α greater than α This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the population mean is not equal to 56.1. There is not sufficient evidence to warrant rejection of the claim that the population mean is not
Transcript text: You wish to test the following claim ( $H_{1}$ ) at a significance level of $\alpha=0.05$. \[ \begin{array}{l} H_{0}: \mu=56.1 \\ H_{1}: \mu \neq 56.1 \end{array} \] You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size $n=21$ with mean $\bar{x}=61.8$ and a standard deviation of $s=6.2$. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic $=$ What is the p -value for this sample? (Report answer accurate to four decimal places.) $p$-value $=$ The p-value is... less than (or equal to) $\alpha$ greater than $\alpha$ This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the population mean is not equal to 56.1 . There is not sufficient evidence to warrant rejection of the claim that the population mean is not
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Solution

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Solution Steps

Step 1: Calculate the Standard Error

The standard error \( SE \) is calculated using the formula: \[ SE = \frac{s}{\sqrt{n}} = \frac{6.2}{\sqrt{21}} \approx 1.353 \]

Step 2: Calculate the Test Statistic

The test statistic \( t_{\text{test}} \) is calculated using the formula: \[ t_{\text{test}} = \frac{\bar{x} - \mu_0}{SE} = \frac{61.8 - 56.1}{1.353} \approx 4.213 \]

Step 3: Calculate the P-value

For a two-tailed test, the p-value \( P \) is calculated as: \[ P = 2 \times (1 - T(|z|)) \approx 0.0004 \]

Step 4: Decision Based on P-value

Since the p-value \( 0.0004 \) is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis \( H_0 \).

Step 5: Conclusion

There is sufficient evidence to warrant rejection of the claim that the population mean is not equal to \( 56.1 \).

Final Answer

\[ \text{Test Statistic: } t_{\text{test}} \approx 4.213 \] \[ \text{P-value: } P \approx 0.0004 \] \[ \text{Decision: } \text{reject the null} \] \[ \text{Conclusion: } \text{There is sufficient evidence to warrant rejection of the claim that the population mean is not equal to } 56.1. \] \[ \boxed{t_{\text{test}} \approx 4.213, P \approx 0.0004} \]

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