Question #339396

A researcher claims that 10 year olds watch 6.6 hours of TV daily with SD = 2.5 hours. You try to verify this with the following sample data of 100 and a sample mean of 6.1 hours. Test the claim of the researcher. Test at α = .01




1
Expert's answer
2022-05-11T09:41:51-0400

The following null and alternative hypotheses need to be tested:

H0:μ=6.6H_0:\mu=6.6

H1:μ6.6H_1:\mu\not=6.6

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 is zc=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=6.16.62.5/100=2z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{6.1-6.6}{2.5/\sqrt{100}}=-2

Since it is observed that z=2<2.5758=zc,|z| = 2<2.5758=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<2)=0.0455,p=2P(Z<-2)=0.0455, and since p=0.0455>0.01=α,p = 0.0455 > 0.01=\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 6.6, at the α=0.01\alpha = 0.01 significance level.



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