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On the basis of her newly developed technique, a student believes she can reduce the amount...

On the basis of her newly developed technique, a student believes she can reduce the amount of time schizophrenics spend in an institution. As director of training at a nearby institution, you agree to let her try her method on 20 schizophrenics, randomly sampled from your institution. The mean duration that schizophrenics stay at your institution is 85 weeks, with a standard deviation of 15 weeks. The scores are normally distributed. The results of the experiment show that patients treated by the student stay at the institution a mean duration of 78 weeks. What do you conclude about the student’s technique? Use α = .05.

***Write out all steps of hypothesis testing including populations, hypotheses, cutoff scores, and all relevant calculations.

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Answer #1

The provided sample mean is and the known population standard deviation is , and the sample size is n = 20

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:

Ha:

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation, will be used. The population are the schizophrenics of an institution.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a left-tailed test is .

The rejection region for this left-tailed test is R={z:z<−1.64}

(3) Test Statistics

The z-statistic is computed as follows:

(4) The decision about the null hypothesis

Since it is observed that z=−2.087<zc​=−1.64, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0184, and since p=0.0184<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is less than 85, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval is 71.426<μ<84.574.

Graphically

Let me know in the comments if anything is not clear. I will reply ASAP! Please do upvote if satisfied!

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