22 patients in a treatment group who ate their lunch while playing solitaire wer

Question

22 patients in a treatment group who ate their lunch while playing solitaire were asked to do a serial-order recall of the food lunch items they ate. The average number of items recalled by the patients in this group was 4.9, with a standard deviation of 1.8. The average number of items recalled by the patients in the control group (no distraction) was 6.1, with a standard deviation of 1.8. Do these data provide strong evidence that the average number of food items recalled by the patients in the treatment and control groups are different?

Explanation / Answer

H0: mu1 =mu2

Ha: m1 not equals mu2

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z for 95% CI= 1.96
declare p larger than alpha=0.05 not significant.

mean1 eq: 4.9 (variance= 3.24) (se= 0.3838)
mean2 eq: 6.1 (variance= 3.24) (se= 0.3838)

Probability that var1<var2
p=0.5 (left: 0.5; double: 1)

Difference between means:
M1-M2=4.9-6.1=-1.2
sd=3.5172; se=0.5427
95% CI of difference:
-2.2637 <-1.2< -0.1363
t-difference: -2.211
df-t: 41.5; p=
; two sided: 0.0326)

As p = 0.0326<0.05 reject null hypothesis.

The data provides strong evidence that the average number of food items recalled by the patients in the treatment and control groups are different

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