Aresearcher conducts a repeated-measures study to evaluate a treatment with a sample of n = 16 participants and obtains a t statistic of t = 1.94. the treatment is expected to increase scores and the sample mean shows an increase. what is the correct decision for a hypothesis test using α = .05?

Given : Sample size : n= 16

Degree of freedom = n-1=15

The obtained t-statistic value = 1.94

Since, The treatment is expected to increase scores and the sample mean shows an increase.

Let be the population mean before and denotes the population mean after the treatment.

then the related hypothesis will be :-

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

The critical value for =1.753

Since, the obtained value (1.94) is greater than the critical value (1.753) so we reject the null hypothesis  .

Therefore, we have enough evidence to support the alternative hypothesis.

Hence, we conclude that treatment may successful to increase scores and the sample mean shows an increase.

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

1) is right

step-by-step explanation: