1. To evaluate the effect of a treatment, a sample (n = 16) is obtained from a population with a mean of µ = 30 and a treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 31.3 with a sample standard deviation of s = 3. Are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with α = .05?

We are given that to evaluate the effect of a treatment, a sample (n = 16) is obtained from a population with a mean of µ = 30 and a treatment is administered to the individuals in the sample.

After treatment, the sample mean is found to be M = 31.3 with a sample standard deviation of s = 3.

Let [tex]\mu[/tex] = population mean.

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 30 {means that the treatment does not has a significant effect}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] [tex]\neq[/tex] 30 {means that the treatment has a significant effect}

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

T.S. = [tex]\frac{M-\mu}{\frac{s}{\sqrt{n}}}[/tex] ~ [tex]t_n_-_1[/tex]

where, M = sample mean = 31.3

s = sample standard deviation = 3

n = sample size = 16

So, test statistics = [tex]\frac{31.3-30}{\frac{3}{\sqrt{16}}}[/tex] ~ [tex]t_1_5[/tex]

= 1.733

The value of t test statistics is 1.733.

Now, at 0.05 significance level the t table gives critical values of -2.131 and 2.131 at 15 degree of freedom for two-tailed test. Since our test statistics lie within the range of critical values of t, 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 treatment does not has a significant effect.