Question

Weinstein, McDermott, and Roediger report that students who were givne questions to be answered while studying new material had better scores when tested on the material compared to students who were simply given an opportunity to reread the material. In a similiar study, an instructor in a large psychology class gave one group of students questions to be answered while studying for the final exam. The overall average for the exam was mean = 73.4, but the n = 16 students who answered questions had a mean of M =78.3 with a standard deviation of s=8.4. For this study, did answering questions while studying produce significantly higher exam scores? Use a one tailed test with alpha = .01.

Answer 1

The standard error value is 2.100

t- statistic value 2.333

Result indicate that, failure to reject the null hypothesis; answering questions while studying did not produce significantly higher exam scores. Option d is right choice.

Answering questions during study time does not significantly alter exam scores, according to the null hypothesis (
`H_O`) (i.e., there is no difference in means between the group that answered questions and the overall average).

The alternative hypothesis (
`H_1`) states that exam scores are considerably higher when study questions are answered.

To find the standard error use the formula and put the given value

standard error =
`S_m` =
`(s)/(√(n) ) = (8.4)/(√(16) ) = 2.100`

0.05 level with two tail test and n-1= 15 df, critical t = 2.131

Now for t- statistic

t- statistic =
`('(\bar x - \mu))/(sx)`

t- statistic =
`(78.3 - 73.4)/(2.1)`

t- statistic = 2.333

The computed t-value is within the critical t-value range, hence the null hypothesis cannot be ruled out.

Cohen's d =
`(x-\mu_0)/(s) = (\rho)/(\sigma) =0.583`

variance accounted =
`r^2` =
`(t^2)/(t^2 +df) = 0.266`

It shws that, failure to reject the null hypothesis; answering questions while studying did not produce significantly higher exam scores.

Option d is right choice.

Answer 2

Answer with explanation:

Let
`\mu` be the population mean.

By considering the given information , we have

Null hypothesis :
`H_0: \mu=73.4`

Alternative hypothesis :
`H_a: \mu>73.4`

Since alternative hypothesis is right-tailed , so the test is a right-tailed test.

Given : Sample size : n=16 , which is a small sample , so we use t-test.

Sample mean:
`\overline{x}=78.3` ;

Standard deviation:
`s=8.4`

Test statistic for population mean:

`t=\frac{\overline{x}-\mu}{(s)/(√(n))}`

i.e.
`t=(78.3-73.4)/((8.4)/(√(16)))\approx2.333`

Using the standard normal distribution table of t , we have

Critical value for
`\alpha=0.01` :
`t_((n-1,\alpha))=t_((15,0.01))=2.602`

Since , the absolute value of t (2.333) is smaller than the critical value of t (2.602) , it means we do not have sufficient evidence to reject the null hypothesis.

Hence, we conclude that we do not have enough evidence to support the claim that answering questions while studying produce significantly higher exam scores.