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For each problem students will write out all steps of hypothesis testing including populations, hypotheses, cutoff scores, and all relevant calculations. Assignments will be typed and uploaded in a word document to blackboard. 2. 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 students technique? Use a .05. D Focus lish (United States)
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The provided sample mean is X = 85 and the sample standard deviation is s = 15, and the sample size is n = 20. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u 278 Ha: μ < 78 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a 0.05, and the critical value for a left-tailed test is te1.729 The rejection region for this left-tailed test is R t:t1.729

(3) Test Statistics The t-statistic is computed as follows: 85 78 t=-= =2.087 (4) Decision about the null hypothesis Since it is observed that 2.087e.729, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.9747, and since p = 0.9747 > 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected.  Therefore, there is not enough evidence to claim that a student believes she can reduce the amount of time schizophrenics spend in an institute

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