Question #336489

ACTIVITY 8. APPLYING HYPOTHESIS TESTING

Solve the problem:

A recent survey says that Filipino children spend an average of 4 hr a day playing online games with a standard deviation of 30 minutes. A random sample of 9 children is taken from a normally distributed population of children who spend an average of 4 hr and 30 min playing online games. Using the 1% level of significance, would you conclude that the statement given in the survey is correct?


1
Expert's answer
2022-05-03T10:59:08-0400

The following null and alternative hypotheses need to be tested:

H0:μ=4H_0:\mu=4

Ha:μ4H_a:\mu\not=4

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is α=0.01,\alpha = 0.01, and the critical value for a two-tailed test iszc=2.5758.z_c = 2.5758.

The rejection region for this two-tailed test is R={z:z>2.5758}.R = \{z: |z| > 2.5758\}.

The z-statistic is computed as follows:


z=xˉμσ/n=4.540.5/9=3z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{4.5-4}{0.5/\sqrt{9}}=3

Since it is observed that z=3>2.5758=zc,|z| = 3 >2.5758= z_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p=2P(Z>3)=2(0.00135)=0.0027,p =2P(Z>3)=2(0.00135)=0.0027, and since p=0.0027<0.01=α,p = 0.0027 < 0.01=\alpha, it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ\mu  is different than 4, at the α=0.01\alpha = 0.01 significance level.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS