Short Answer

A health psychologist is calculating a one-sample tt-test for a sample of N=10N = 10 participants. The sample mean is M=212M = 212, the hypothetical population mean is μ0=250\mu_0 = 250, and the sample standard deviation is SD=39.17SD = 39.17. Apply the formula t=Mμ0SDNt = \frac{M - \mu_0}{\frac{SD}{\sqrt{N}}} to calculate the value of the tt statistic. Show your intermediate calculations.

Question: A health psychologist is calculating a one-sample tt-test for a sample of N=10N = 10 participants. The sample mean is M=212M = 212, the hypothetical population mean is μ0=250\mu_0 = 250, and the sample standard deviation is SD=39.17SD = 39.17. Apply the formula t=Mμ0SDNt = \frac{M - \mu_0}{\frac{SD}{\sqrt{N}}} to calculate the value of the tt statistic. Show your intermediate calculations.

Sample answer: To calculate the tt statistic: 1) Calculate the denominator: SDN=39.171012.39\frac{SD}{\sqrt{N}} = \frac{39.17}{\sqrt{10}} \approx 12.39. 2) Calculate the numerator: Mμ0=212250=38M - \mu_0 = 212 - 250 = -38. 3) Calculate the final tt statistic: t=3812.393.07t = \frac{-38}{12.39} \approx -3.07.

Key points:

  • State the calculation of the numerator (212250=38212 - 250 = -38)
  • State the calculation of the denominator (39.171012.39\frac{39.17}{\sqrt{10}} \approx 12.39)
  • Calculate the final tt statistic value (3.07-3.07)

Rubric: Full credit is awarded if the student correctly applies the formula and obtains a final tt value of 3.07-3.07 (or close rounding), showing the numerator of 38-38 and denominator of approximately 12.3912.39.

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Updated 2026-05-27

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Research Methods in Psychology - 4th American Edition @ KPU

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