Question

An educational psychologist wants to test whether a new teaching method negatively affects reading comprehension scores. She randomly selects 30 6th grade students that were taught under the new teaching method and finds that their scores on a standardized reading comprehension test have a mean equal to 118.8 with a variance equal to 37.2. Scores on the standardized test in the general population of 6th graders are distributed approximately normally with a mean equal to 119.8. Is there sufficient evidence to conclude that the new teaching method negatively affects reading comprehension scores

Answer 1

Answer:

We conclude that the new teaching method does not negatively affects reading comprehension scores.

Step-by-step explanation:

We are given that an educational psychologist wants to test whether a new teaching method negatively affects reading comprehension scores.

She randomly selects 30 6th grade students that were taught under the new teaching method and finds that their scores on a standardized reading comprehension test have a mean equal to 118.8 with a variance equal to 37.2.

Scores on the standardized test in the general population of 6th graders are distributed approximately normally with a mean equal to 119.8.

Let $\mu$ = mean scores on a standardized reading comprehension test.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 119.8      {means that the new teaching method does not negatively affects reading comprehension scores}

Alternate Hypothesis, $H_A$ : $\mu$ < 119.8    {means that the new teaching method negatively affects reading comprehension scores}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

                      T.S. =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean test score = 118.8

            s = sample standard deviation = $\sqrt{37.2}$ = 6.1

            n = sample of 6th grade students = 30

So, test statistics  =  $\frac{118.8-119.8}{\frac{6.1}{\sqrt{30} } }$  ~ $t_2_9$

                              =  -0.898

The value of t test statistics is -0.898.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical value of -1.699 for left-tailed test.

Since our test statistic is more than the critical value of t as -0.898 > -1.699, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the new teaching method does not negatively affects reading comprehension scores.