Question

Dr. Calvin Broadus wants to determine whether marijuana influences anxiety. He draws a sample of n = 16 individuals. The population from which they were selected exhibit a score of µ = 30 on an anxiety measure. After “treatment” is administered to the individuals, the sample mean is found to be M = 33 and the sample variance is s² = 64.

Is this a one-tailed or two-tailed test?

		One-tailed
		Two-tailed

If the researcher decides to test at the α = .05 level, what are degrees of freedom and t_critical value?

		df = 15,  tcritical =   ± 2.131
		df = 16,  tcritical =  ± 2.120
		df = 15,  tcritical =  + 1.753
		df = 16,  tcritical =  + 1.746

In determining whether the data are sufficient to conclude if marijuana affects anxiety, what is the observed test statistic value?

		t = -0.38
		t = 1.50
		t = 3.00
		t = -0.75

Based on the observed test statistics, would Dr. Broadus reject or retain the null hypothesis?

		Reject Null Hypothesis (H0)
		Retain Null Hypothesis (H0)

What would Dr. Broadus then conclude about the effect of marijuana on anxiety?

		There is insufficient evidence to determine whether marijuana influences anxiety.
		Marijuana significantly increases anxiety scores.

Construct a 95% confidence interval for an estimate of the population mean.

		95% CI = 29.48 to 36.52
		95% CI = 25.74 to 33.52
		95% CI = 28.74 to 37.26
		95% CI = 30.00 to 33.00

Calculate the effect size in terms of proportion of the variance account for (r²)

		r2 = .75
		r2 = .38
		r2 = .13
		r2 = -.13

Answer 1

1. Since whether marijuana influences anxiety is being tested, we have the null hypothesis

H0: mean= 30

and alternative hypothesis

H1: mean is not equal to 30.

Hence this is a two tailed test.

2. the t Statistic used here follows a t distribution with (n-1)= 16-1 = 15 degrees of freedom and the critical value of t for 15 degrees of freedom and alpha= 0.05 for a two-tailed test is given by 2.131

3. The t Statistic is given by

t= (xbar- mean)/ S/sqrt(n)

= (33-30)/8/sqrt(16)

= 3/8/4

= 12/8 = 1.50

4. Since calculated t= 1.50< critical t= 2.33

Hence, H0 can't be rejected, thus H0 will be retained.

5. Since H0 is not rejected, there is insufficient evidence to determine whether marijuana influences anxiety.