Question

A random sample of 150 workers with children in day care shows a mean day care cost Rs. 3200 and a standard deviation of Rs. 750. Verify the department’s claim that the mean does not exceeds Rs. 3000 at 5% level of significance with this information.

Answer 1

z = (3200-3000)/(750/$\sqrt{150}$ ) = 3.27

P(z>3.27)=0.000538.

 Mean does exceed Rs. 3000.

The result is significant at 5% level of significance.