Question

A school counselor was interested in the effect of a new program designed to reduce adjustment problems in newly transferred students. From years of experience working at the school, the counselor knew that the average score on a scale of adjustment difficulties for a transfer student was 58 with a standard deviation of 10. After the new program had been implemented, the adjustment of 50 transfer students was measured and their mean score was 52. Was this difference significant at the .01 level? a. Use the five steps of hypothesis testing. b. Figure the confidence limits for the 99% confidence interval.

Answer 1

a) i) First, we would assume that the data on adjustment difficulty score is normally distributed and the 50 students were randomly chosen.

ii) The null and the alternate hypothesis are

$H_{o} : \mu = 58$

$H_{a} : \mu < 58$

iii) The average score for 50 students is also normally distributed with mean = 58 and the standard error $= \sigma/\sqrt{n}$ .

So the sampling distribution of $\overline{x}$ is $\sim N(\mu, \sigma^2/n )$ .

iv) The value of the test statistics is

$z = \frac{\overline{x} - \mu}{\sigma/\sqrt{n}}$

$z = \frac{52 - 58}{10/\sqrt{50}} = -4.243$

v) The p-value corresponding to the test statistics is = 0.000011.

The significance level is = 0.01.

Since the p-value is less than the significance level, hence we will reject the null.

Conclusion: The new program designed to reduce the adjustment problem in newly transferred students is effective.

b) The confidence limit for the 99% confidence interval is

$= \mu - z_{\alpha} * \sigma/\sqrt{n}$ (There is only a lower limit since it is a left-tailed test. So , if the average value is less than this lower limit then we will reject the null hypothesis).

$= 58 - 2.33 * 2.2/\sqrt{50}$

As we can again see, the average value of 52 is less than the lower limit calculated above hence we will reject the nul hypothesis.

Thank You!!

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