In November 2024
Answer to Question #187814 in Statistics and Probability for tahreem anwar
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.
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.
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#339914 on Dec 2023
Comments
Dear MUHAMMAD HASSAM EJAZ, a solution of the question was published. Please describe in detail issues you observed in a solution of the question.
That's a good way of solving this tricky paper, but I don't think anyone is going to reply.
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