Question #334540

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.

1
Expert's answer
2022-04-28T15:55:41-0400

The following null and alternative hypotheses need to be tested:

H0:μ14200H_0:\mu\le14200

Ha:μ>14200H_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 α=0.10,\alpha = 0.10, df=n1=201=19df=n-1=20-1=19 degrees of freedom, and the critical value for a right-tailed test is tc=1.327728t_c = 1.327728

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

The t-statistic is computed as follows:


t=xˉμs/n=131002.1864142002250/202.1864t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{131002.1864-14200}{2250/\sqrt{20}}\approx-2.1864

Since it is observed that t=2.18641.327728=tc,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=19df=19 degrees of freedom, t=2.1864t=-2.1864 is p=0.979249,p = 0.979249, and since p=0.979249>0.10=α,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 μ\mu

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



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