Question

In a study on the effects stress on behaviour in rats, 71 rats were randomly assigned to either a stressful environment or a control (non-stressful) environment.
After 21 days, the change in weight (in grams) was determined for each rat.

The following table summarises the data on weight gain:

	n	sample mean	sample standard deviation
Stress	21	26.03	12.09
No stress	50	31.61	12.24

1.1 (.8 marks)

The psychologist who conducted the experiment wants to conduct a t-test to determine which condition is going to lead to a greater weight gain, but has no idea beforehand which one. How should the null and alternative hypotheses be framed?

Calculate the value of the t-statistic to three decimal places.
t ≈  

Using the simple method, how many degrees of freedom df does this t-statistic have? df =

Calculate the p-value to three decimal places. P-value ≈

The psychologist is willing to reject the null hypothesis if the p-value is less than α = 0.05. Should the psychologist reject the null hypothesis?

Now we require you to calculate a (1 − α) ×100 = 95% confidence interval for the true mean difference between μ₁ and μ₂.

In order to do that, what, to three decimal places, is the value of the t-statistic t* that you should be using in the expression for the confidence interval? t* ≈

Enter the confidence interval (lo, hi), where lo and hi are the lower and upper limits, respectively, to three decimal places.

(lo, hi) = (

)

Is the result from the hypothesis test consistent with the confidence interval?

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05 Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 3.15543
DF = 69
t = [ (x₁ - x₂) - d ] / SE

t = - 1.77

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 69 degrees of freedom is more extreme than -1.77; that is, less than -1.77 or greater than 1.77.

Thus, the P-value = 0.081

Interpret results. Since the P-value (0.081) is greater than the significance level (0.05), we have to accept the null hypothesis.

D.F = 69

α = 0.05

t_critical = + 1.995

C.I = (26.03 - 31.61) + 1.995*3.15543

C.I = - 5.58 + 6.2951

C.I = (- 11.875, - 0.7151)

Yes, the the result from the hypothesis test consistent with the confidence interval, becuase the confidence interval contains 0.