Question

A random sample of n - 16 scores is selecdted from a normal population with a mean of p - 50. After atreatment is administered to the individuals in the sample, the sample mean is found to be M -54 If the population standard deviation is σ-8, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with α-.05. (Hint: Recall that the critical value for a two-tailed test with α-.05 is z <-1.96 or z > 1.96.) with a 8,2 = 4.00 . You conclude that the treatment has a significant effect. can If the population standard deviation is σ-12, is the sample mean sufficient to conclude that the treatmenthas significant effect? Use a two-tailed test with α-,05. with σ 12, z- . You conclude that the treatment has a significant effect. Comparing your answers for the two cases you just tested, explain how the magnitude of the standard deviation influences the outcome of a hypothesis test. A larger standard deviation the likelihood of rejecting the null hypothesis. ※ Graded

Answer 1

when, $\sigma$ = 8

$z = \frac{M - \mu}{\sigma / \sqrt n} = \frac{54 - 50}{8 / \sqrt {16}} = 2$

As 2 > 1.96, we can conclude that the treatment has a significant effect. So we reject the null hypothesis.

when, $\sigma$ = 12

$z = \frac{M - \mu}{\sigma / \sqrt n} = \frac{54 - 50}{12 / \sqrt {16}} = 1.33$

As 1.33 < 1.96, we can not conclude that the treatment has a significant effect. So we fail to reject the null hypothesis.

As the magnitude of standard deviation increases the chance of rejecting null hypothesis reduces. We saw in our test that as $\sigma$ changes from 8 to 12, the hypothesis which was rejected earlier became fail to reject as the $\sigma$ increases. A larger standard deviation is not the likelihood of rejecting null hypothesis.

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