Answer to Question #187900 in Statistics and Probability for Osama

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. Explain your results in your own words.

Answer 1

Let the population mean be $\mu$ = Rs. 3000

Let the population standard deviation be $\delta$ = Rs. 750

We are testing the following alternative hypothesis against its corresponding null hypothesis.

H₀: $\mu$ = Rs. 3000

H₁: $\mu$ > Rs. 3000

The provided $\alpha$ = 0.05

We will therefore conduct a 1-tailed test.

Using the standard normal table, we determine the critical z-score at $\alpha$ = 0.05

The z score (z=1.645) shows the edge of the rejection region.

To find out the computed z-score, we use the following formula

$z=\frac{( x̄ - μ)}{\frac{\sigma}{\sqrt{n}}}$

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

We observe that the computed z-score from the data is greater than the given z score of z=1.645

The researchers therefore reject the null hypothesis in favor of the alternative hypothesis, and conclude that there is enough evidence to support the claim that the mean day care cost exceed

Rs. 3000.