Answer to Question #128786 in Statistics and Probability for Ahtisham

Question

A random sample of 100 workers with childrens in day care show a mean day care cost of Rs. 2600 and a standard deviation of Rs. 500.verify the department claim that the mean exceeds Rs. 2500 at the 0.05 level with this information

Answer 1

Let the population mean be $\mu$ = Rs. 2500

Let the population standard deviation be $\delta$ = Rs. 500

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

H₀: $\mu$ = Rs. 2500

H₁: $\mu$ > Rs. 2500

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

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

$z=\frac{(2600-2500)}{\frac{500}{\sqrt{100}}}$ = 2

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. 2500.