Performing a One-Sample $t$-Test

We are conducting a one-sample $t$-test to determine whether the sample mean ($\bar{x}$) is significantly different from the population mean ($\mu_{\text{FPL}}$).

Given:

• $\mu_{\text{FPL}} = 15,060$ (population mean)
• $\bar{x} = 7,151.88$ (sample mean)
• $n = 16$ (sample size)
• $s = 1,250.35$ (sample standard deviation)
• $\alpha$: Not specified (commonly $0.05$)

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Steps:

1. Formulate hypotheses:

   • Null Hypothesis ($H_0$): $\mu = 15,060$ (no significant difference)
   • Alternative Hypothesis ($H_a$): $\mu \neq 15,060$ (there is a significant difference)

2. Calculate the test statistic ($t$): The formula for $t$ is: $t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$

   Substituting the values: $t = \frac{7,151.88 - 15,060}{1,250.35 / \sqrt{16}}$ $t = \frac{-7,908.12}{312.5875} = -25.29$

3. Degrees of freedom: $df = n - 1 = 16 - 1 = 15$

4. Determine the critical $t$-value: Using $df = 15$ and a two-tailed test with $\alpha = 0.05$, the critical $t$-value is approximately $\pm 2.131$.

5. Compare $t$-value to critical value: $t = -25.29$ Since $t$ is far less than $-2.131$, we reject $H_0$.

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Conclusion:

The $t$-statistic of $-25.29$ falls outside the critical range ($\pm 2.131$). Therefore, we reject the null hypothesis and conclude that the sample mean ($7,151.88$) is significantly different from the population mean ($15,060$).

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Would you like me to calculate the $p$-value or provide more details about the interpretation? 😊

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Related Questions:

1. How is the $t$-distribution used for small sample sizes?
2. Why do we divide by $\sqrt{n}$ in the $t$-test formula?
3. What assumptions are required for a one-sample $t$-test?
4. How does the sample size affect the $t$-statistic?
5. What is the difference between one-tailed and two-tailed $t$-tests?

Tip:

When performing a $t$-test, always verify the assumptions (e.g., normality of data or large enough $n$) to ensure the results are valid.