Answer to Question #334540 in Statistics and Probability for Kenneth

Question

The average pre-school cost for tuition fees last year was ₱14,200. The following year, 20

institutions had a mean of ₱13,100 and a standard deviation of ₱2,250. Is there sufficient evidence

at 𝛼= 0.10 level of significance to conclude that the mean cost has increased?

SOLUTION:

Step 1 𝐻0:____________________________________________________

𝐻𝑎: ____________________________________________________

Step 2. 𝛼= _________

Step 3. 𝑧𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 = _________

Step 4. Compute the test statistic.

Step 5. Decision Rule:_______________________

Step 6: Conclusion.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le14200$

$H_a:\mu>14200$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.10,$ $df=n-1=20-1=19$ degrees of freedom, and the critical value for a right-tailed test is $t_c = 1.327728$

The rejection region for this right-tailed test is $R = \{t: t > 1.327728\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{131002.1864-14200}{2250/\sqrt{20}}\approx-2.1864$$

Since it is observed that $t = -2.1864 \le1.327728=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, $df=19$ degrees of freedom, $t=-2.1864$ is $p = 0.979249,$ and since $p=0.979249>0.10=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

s greater than 14200, at the $\alpha = 0.10$ significance level.