Question #350962

A medical centre reports that the average cost of rehabilitation for stroke victims is 24672 BD. To see if the average cost of rehabilitation is different at a particular hospital, a researcher selects 35 victims at random and find the average cost is 26343 BD. Standard deviation is 3251 BD. At α=0.01, can it be concluded that the average cost of rehabilitation at a particular hospital is different form 24672 BD


1
Expert's answer
2022-06-16T10:27:32-0400

The following null and alternative hypotheses need to be tested:

H0:μ=24672H_0:\mu=24672

H1:μ24672H_1:\mu\not=24672

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is α=0.01,\alpha = 0.01, df=n1=34df=n-1=34 and the critical value for a two-tailed test is tc=2.728394.t_c =2.728394.

The rejection region for this two-tailed test is R={t:t>2.728394}.R = \{t:|t|>2.728394\}.

The t-statistic is computed as follows:


t=xˉμs/n=26343246723251/35=3.0408t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{26343-24672}{3251/\sqrt{35}}=3.0408


Since it is observed that t=3.0408>2.728394=tc,|t|=3.0408>2.728394=t_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, df=34df=34 degrees of freedom, t=3.0408t=3.0408 is p=0.004521,p= 0.004521, and since p=0.004521<0.01=α,p=0.004521<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 different than 24672, at the α=0.01\alpha = 0.01 significance level.


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