In: Statistics and Probability
In a study of memory recall, ten students from a large
statistics and data analysis class were
selected at random and given 15 minutes to memorize a list of 20
nonsense words. Each was
asked to list as many of the words as he or she could remember both
1 hour and 24 hours later.
The data are as shown in Table 1. Is there evidence to suggest the
mean number of words recalled
after 1 hour exceeds the mean recall after 24 hours by more than 3?
Use a significance level of
0.05.
students | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
1 hours later (words) | 14 | 12 | 18 | 7 | 11 | 9 | 16 | 15 | 13 | 17 |
24 hours later(words) | 10 | 4 | 14 | 5 | 8 | 5 | 11 | 12 | 9 | 10 |
Solution :
Null and alternative hypotheses :
The null and alternative hypotheses would be as follows :
Test statistic :
To test the hypothesis the most appropriate test is paired t-test. The test statistic is given as follows :
Where, is sample mean of the differences, is hypothesized mean difference under H0, s is sample standard deviation of differences and n is sample size.
From the above table we have,
The value of the test statistic is 1.4.
P-value :
Since, our test is right-tailed test, therefore we shall obtain the right-tailed p-value for the test statistic. The right-tailed p-value is given as follows :
P-value = P(T > t)
P-value = P(T > 1.4)
P-value = 0.1102
The p-value is 0.1102.
Decision :
Significance level = 0.05
(0.1102 > 0.05)
Since, p-value is greater than the significance level of 0.05, therefore we shall be fail to reject the null hypothesis (H0) at 0.05 significance level.
Conclusion :
At 0.05 significance level, there is not sufficient evidence to suggest the mean number of words recalled after 1 hour exceeds the mean recall after 24 hours by more than 3.
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