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=M−μ0SDNt = \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=M−μ0SDNt = \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=39.1710≈12.39\frac{SD}{\sqrt{N}} = \frac{39.17}{\sqrt{10}} \approx 12.39. 2) Calculate the numerator: M−μ0=212−250=−38M - \mu_0 = 212 - 250 = -38. 3) Calculate the final tt statistic: t=−3812.39≈−3.07t = \frac{-38}{12.39} \approx -3.07.

Key points:

  • State the calculation of the numerator (212−250=−38212 - 250 = -38)
  • State the calculation of the denominator (39.1710≈12.39\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 12.39.

0

1

Updated 2026-05-27

Contributors are:

Who are from:

Tags

KPU

Research Methods in Psychology - 4th American Edition @ KPU

Related