Question #340704

A mathematics teacher in senior high school developed a problem-solving test to randomly selected 40 grade 11 students. These students had an average score of 85 and a standard deviation of 5. If the population had a mean score of 90 and a standard deviation of 3, use 5% level of significance to test the hypothesis.


1
Expert's answer
2022-05-15T23:12:16-0400

The following null and alternative hypotheses need to be tested:

H0:μ=90H_0:\mu=90

H1:μ90H_1:\mu\not=90

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=85903/4010.541z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{85-90}{3/\sqrt{40}}\approx-10.541

Since it is observed that z=10.541>1.96=zc,|z| = 10.541 > 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<10.541)0,p=2P(z<-10.541)\approx0, and since p=0<0.05=α,p=0<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 90, at the α=0.05\alpha = 0.05 significance level.


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