Question #346547

1. The average score in the entrance examination in Mathematics at Sto.



Rosario National High School is 80 with a standard deviation of 10.A



random of 40 students was taken from this year's examinees and



was found to have a mean score of 84.



Is there a significant difference between the known mean and the



sample mean? Test at a= 0.05



Solution :



Step 1. H0: = 80 : There is no significant difference



hypothesized and the sample mean.



H1: \neq 80 :



Step 2. Level of significance. a = 0.05.



Step 3. two tailed test, find the critical value. Zt =



4. Compute the test-statistic value:


5. Step :

6. Conclusion:


1
Expert's answer
2022-06-01T14:24:24-0400

The following null and alternative hypotheses need to be tested:

H0:μ=80H_0:\mu=80

H1:μ80H_1:\mu\not=80

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=0.05,\alpha = 0.05, and the critical value for a two-tailed test is zc=1.96.z_c = 1.96.

The rejection region for this two-tailed test is R={z:z>1.96}.R = \{z:|z|>1.96\}.

The z-statistic is computed as follows:



z=xˉμσ/n=848010/40=2.5298z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{84-80}{10/\sqrt{40}}=2.5298

Since it is observed that z=2.5298>1.96=zc,|z|=2.5298>1.96=z_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p=2P(z>2.5298)=0.011413,p=2P(z>2.5298)=0.011413, and since p=0.011413<0.05=α,p= 0.011413<0.05=\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 80, at the α=0.05\alpha = 0.05 significance level.


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