Question

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.

Answer 1

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.