A study is conducted to determine if a newly designed text book is more helpful

Question

A study is conducted to determine if a newly designed text book is more helpful to learning the material than the old edition. The mean score on the final exam for a course using the old edition is 75. Scores follow a normal distribution. Ten randomly selected people who used the new text take the final exam. Their scores are shown in the table below. Person A B C D E F G H I J Test Score 94 96 87 92 70 75 84 80 74 68 Use a 0.01 significance level to test the claim that people do better with the new edition. (1.) What kind of test should be used? A. Two-Tailed B. One-Tailed C. It does not matter.

(2.) The test statistic is (rounded to 2 decimals).

(3.) The P-value is

(4.) Is there sufficient evidence to support the claim that people do better than 75 on this exam? A. Yes B. No

(5.) Construct a 99% confidence interval for the mean score for students using the new text.

Explanation / Answer

mu = 75

and X the mean score is normal

n = sample size = 10

Since sample size is <30, t test can be used

H0: mu = 75

Ha: mu >75

Hence one tailed test.

The sample given has mean x bar = 81.7 and

std dev = s =9.787

Std error = s/rt n-1 = 3.262

Mean difference = 6.7

Test stat = 6.7/3.262 = 2.054

p value =.035079.

The result is not significant at p < .01.

i.e. there is no  sufficient evidence to support the claim that people do better than 75 on this exam

No.

5) 99% conf interval = Mean + 2.8214*3.262

= 75+9.2034

= (75, 84.2034)

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