A random sample of 150workers with children in day care shows a mean dry care of 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.
Let "H_o" : Man does not exceed Rs 3000 at 5% level of significance
"\\mu=3200, \\sigma=750"
"\\alpha=0.05"
Critical value -
"z_{\\alpha}=z_{0.05}=1.645"
"P(x\\le 3000)=P(z<\\dfrac{3000-3200}{\\frac{750}{\\sqrt{150}}})=P(z<-3.265)=0.000947"
Conclusion: As Calculated value of z is less than the critical value 1.645, So "H_o" is accepted. There is enough evidence to support the department's claim.
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