18. a. A repeated-measures study with a sample of n = 16 participants produces a mean difference of MD = 4 with a standard deviation of s = 8. Use a two-tailed hypothesis test with α = .05 to determine whether it is likely that this sample came from a population with µD = 0.b. Now assume that the sample mean difference is MD = 10, and once again visualize the sample distribution. Use a two-tailed hypothesis test with α = .05 to determine whether it is likely that this sample came from a population with µD = 0.c. Explain how the size of the sample mean difference influences the likelihood

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Asked Jan 16, 2020
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18. a. A repeated-measures study with a sample of n = 16 participants produces a mean difference of MD = 4 with a standard deviation of s = 8. Use a two-tailed hypothesis test with α = .05 to determine whether it is likely that this sample came from a population with µD = 0.

b. Now assume that the sample mean difference is MD = 10, and once again visualize the sample distribution. Use a two-tailed hypothesis test with α = .05 to determine whether it is likely that this sample came from a population with µD = 0.

c. Explain how the size of the sample mean difference influences the likelihood

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Step 1

a)

Null hypothesis:

µD=0.

Alternative hypothesis:

µD0

This is a two tailed test.

Since the population standard deviation is unknown, the appropriate test is one sample t-test.

The test statistic value can be obtained as follows:

Step 2

Computation of P-value:

The P-value for the two tailed t-value at 15 degrees of freedom is 0.0639 which can be can be obtained using the excel formula “=T.DIST.2T (2, 14)”.

Assume that the level of significance is 0.05.

Decision rule:

If p-value ≤α, then reject the null hypothesis. Otherwise, do not reject the null hypothesis.

Conclusion:

Here, p-value (0.0639) is greater than the level of significance (0.05).

Therefore, fail reject the null hypothesis.

There is sufficient evidence to conclude that the sample is came from a population with µD=0.

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