Asked Jan 15, 2020

10. In 1974, Loftus and Palmer conducted a classic study demonstrating how the language used to ask a question can influence eyewitness memory. In the study, college students watched a film of an automobile accident and then were asked questions about what they saw. One group was asked, “About how fast were the cars going when they smashed into each other?” Another group was asked the same question except the verb was changed to “hit” instead of “smashed into.” The “smashed into” group reported significantly higher estimates of speed than the “hit” group. Suppose a researcher repeats this study with a sample of today’s college students and obtains the following results.

a. Do the results indicate a significantly higher estimated speed for the “smashed into” group? Use a one-tailed test with α = .01.

b. Compute the estimated value for Cohen’s d to measure the size of the effect.

c. Write a sentence demonstrating how the results of the hypothesis test and the measure of effect size would appear in a research report.



Estimated Speed
Smashed into
n = 15
n = 15
M = 40.8
M = 34.0
SS = 414
SS = 510

Image Transcriptionclose

Estimated Speed Smashed into Hit n = 15 n = 15 M = 40.8 M = 34.0 %3D SS = 414 SS = 510 %3!


Expert Answer

Step 1


The null and alternative hypotheses are:

H0: The estimated speeds for the “smashed into” and “hit” groups are not different.

H1: The estimated speed for the “smashed into” group is greater than that of the “hit” group.

Step 2

Formulae used:

Denote SS as the sum of squares, Sp2 as the pooled variance, Se as the standard error of estimate.

Statistics homework question answer, step 2, image 1
Step 3

Calculation (a):

From the formula for the pooled variance, it can be understood that SS1 = (n1 – 1) S12, SS2 = (n2 – 1) S22.

Pooled variance:


= (S12 + S22) / (n1 + n2 – 2)

= (510 + 414) / (15 + 15 – 2)

= 924/28

= 33.

Standard error:


= Sp / √ (1/n1 + 1/n2)

= √ [Sp2 / (1/n1 + 1/n2)]

= √ [33/(1/15 + 1/15)]

≈ 2.0976.

Test statistic:


= (1 – 2)/Se

= (40.8 – 34.0)/2.0976

≈ 3.2418.


The degrees of freedom for the t-test is n1 + n2 – 2 = 28.

For the right-tailed test at level of significance 0.01 with 28 degrees of freedom, the t-test cutoff value is 2.47.

Decision rule: Reject H...

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