A local school district claims that the number of school days missed by its teachers due to illness is below the national average of 5 with a population standard deviation of 2.119. A random sample of 40 teachers provided the data below. Test the district's claim using a=0.05. Show graph.
0 | 3 | 6 | 3 | 3 | 5 | 4 | 1 | 3 | 5 |
7 | 3 | 1 | 2 | 3 | 3 | 2 | 4 | 1 | 6 |
2 | 5 | 2 | 8 | 3 | 1 | 2 | 5 | 4 | 1 |
1 | 1 | 2 | 1 | 5 | 7 | 5 | 4 | 9 | 3 |
For the given random sample
mean=sum of total number/ total sample
mean=(0+3+6+3+....4++9+3)/40=3.4
population mean=5
standard deviation of population=2.119
H0 :u>=5
H1:u<5
z=(3.4-5)/[2.119/sqrt(40)]=-4.77
Z at aplha =0.05= should lies between -1.96 and +1.96 from Z table
therefore z calculated i.e -4.77 is less than-1.96 reject the null hypothesis.
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