Question #348114

The principal claims that the students in his school are above average intelligence. A random sample of 35 IQ scores have a mean score of 110.5. Is there sufficient evidence to support the principal’s claim? The mean population IQ is 100 with a standard deviation of 25. Use =1%.


1
Expert's answer
2022-06-07T15:59:15-0400

The following null and alternative hypotheses need to be tested:

H0:μ100H_0:\mu\le100

H1:μ>100H_1:\mu>100

This corresponds to a right-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 right-tailed test is zc=2.3263.z_c = 2.3263.

The rejection region for this right-tailed test is R={z:z>2.3263}.R = \{z:z>2.3263\}.

The z-statistic is computed as follows:


z=xˉμσ/n=110.510025/35=2.4848z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{110.5-100}{25/\sqrt{35}}=2.4848

Since it is observed that z=2.4848>2.3263=zc,z=2.4848>2.3263=z_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p=P(z>2.4848)=0.006481,p=P(z>2.4848)=0.006481, and since p=0.006481<0.01=α,p=0.006481<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 greater than 100, 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