At Hinsdale College, students are employed by the college library to re-shelve books.
The students claim that a 15-minute coffee break every hour would relive boredom and
increase the average number of books re-shelved per hour. To test the student's claim,
the following data were gathered on a random sample of seven student employees.
Over a period of one week, each student worked a normal schedule without coffee
breaks. Then, during the next week, the students were allowed coffee breaks.
Unknown to the students, a member of the library staff secretly recorded the average
number of books shelved per hour and computed the averages with and without the
coffee break system. The results are below. Test the claim that the students shelve
more books when they have a coffee break, use a 0.05 level of significance.
Student 1 2 3 4 5 6 7
Without
Coffee
breaks
75 63 82 53 79 96 73
With
Coffee
Breaks
90 72 95 50 85 110 89

Respuesta :

The p value is less, therefore one should reject the null hypothesis.

How to illustrate the information?

From the information given, the test statistic will be:

t = (10 - 0)/(6.733/✓7)

t = 3.930.

The degree of freedom will be:

= 7 - 1 = 6

Using the table, the p value will be 0.0039.

Since the p value is less than 0.05 one should reject the null hypothesis.

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