Question #298231

The mean Verbal SAT score for the population of first students at Radford is 520. The


standard deviation of scores in this population is 95. An investigator believes that the mean


Verbal SAT of first year psychology majors is significantly different from the mean score of the


population. The mean of a sample of 36 first year psychology majors is 548. Test the


investigator's prediction using an alpha level of .05.

1
Expert's answer
2022-02-16T12:32:02-0500

The following null and alternative hypotheses need to be tested:

H0:μ=520H_0: \mu=520

H1:μ520H_1:\mu\not=520

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=xμσ/n=54852095/36=1.7684z=\dfrac{x-\mu}{\sigma/\sqrt{n}}=\dfrac{548-520}{95/\sqrt{36}}=1.7684

Since it is observed that z=1.7684<1.96=zc,|z|=1.7684<1.96=z_c, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=2P(z<1.7684)=0.076994,p=2P(z<-1.7684)=0.076994, and since p=0.076994>0.05=α,p=0.076994>0.05=\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 is different than 520,520, at the α=0.05\alpha = 0.05 significance level.


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