Answer to Question #131253 in Statistics and Probability for habeel wali

Question

A random sample of 150 workers with children in day care shows a mean day care cost of Rs. 3700 and a standard deviation of Rs. 600. Verify the department claim that the mean is equal to Rs. 3500 at the 0.05 level of significance. (Hint Z0.05 = 1.96) 

Answer 1

Here given:

Accept the Null hypothesis, Department claim if ( -1.96<Z<1.96.

Otherwise Reject the Claim.

Here Given

No. of Workers(n) = 150

Standard deviation( $\sigma$ ) = 600

significance level ($\alpha$ ) = 0.05

Z(0.05) =( -1.96)<Z<1.96

X = 3500

now , Z-value is

Z =( $X-X$ bar)/$\sigma$

=(3500-3700)/600

=(-200)/600

Z = (-1/3)=(-0.33) which is also between (-1.96) to (1.96).

So, the claim is correct. Accept the claim.