Answer to Question #321554 in Statistics and Probability for aecee0514

Question

2. Average Senior High School annual cost of tuition fee for all private schools last year was Php43,700. A random sample of costs this year for 45 private schools indicated that the sample mean was Php45,800 and a sample standard deviation was Php5,600. At 0.01 level of significance, is there sufficient evidence to conclude that the cost increased? Solution: Step 1: State the hypothesis. 𝐻0: _____________________________________________________________ 𝐻1: _____________________________________________________________ Step 2: The level of significance and critical region. 𝛼 = ________ and the 𝑧𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 = __________ Step 3: Compute for the value of one sample t test. 𝑧𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 = _______ Step 4: Decision Rule. ________________________________________________________________ ___________________________________. Step 5: Conclusion: _______________

Answer 1

Step 1: The following null and alternative hypotheses need to be tested:

$H_0:\mu\le43700$

$H_1:\mu>43700$

Step 2: 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 $\alpha = 0.01,$ $df=n-1=44$ and the critical value for a right-tailed test is $t_c =2.414134.$

The rejection region for this right-tailed test is $R = \{t:t>2.414134\}.$

Step 3: The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{45800-43700}{5600/\sqrt{45}}=2.5156$$

Step 4: Since it is observed that $t=2.5156>2.414134=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, $df=44$ degrees of freedom, $t=2.5156$ is $p=0.007804,$ and since $p=0.007804<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Step 5: Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 43700, at the $\alpha = 0.01$ significance level.