Name:______
Assignment #4 Due: Tuesday 12/5
Directions: This entire assignment is to be done by hand and checked using SPSS. Please show your work and attach SPSS output.
1. Each column represents an independent sample of data. For each sample, calculate the mean, variance, standard deviation and perform a one sample t-test. The value of µ, α, and the direction for each test is indicated in the bottom rows. For the one-tailed tests, assume the observed direction of the difference is in the direction hypothesized by the researcher. For each of the two-tailed tests build a 95% confidence interval around the point estimate.
13 / 28 / 20 / 19 / 287 / 23 / 18 / 6 / 27
11 / 23 / 12 / 9 / 27
9 / 23 / 12 / 13 / 23
10 / 26 / 14 / 13 / 23
10 / 26 / 14 / 12 / 24
26 / 15 / 12 / 24
24
µ = 12 / µ = 24 / µ = 20 / µ = 10 / µ = 23
α = .05 / α = .05 / α = .05 / α = .01 / α = .01
1-tailed / 2-tailed / 2-tailed / 1-tailed / 2-tailed
1a. H0 ≥ 12 and H1 < 12
With α = .05 reject H0 if t5 < -2.015
Since a t of –2.449 is less than – 2.015 (p = .058/2 = .029) we reject the null hypothesis.
1b. H0 = 24 and H1 ≠ 24
With α = .05 reject H0 if t6 < -2.447 or > 2.447
One-Sample Statistics
N / Mean / Std. Deviation / Std. Error MeanVAR00002 / 7 / 25.0000 / 2.00000 / .75593
Since a t of 1.323 is greater than – 2.447 and less than 2.447 (p = .234) we do not reject the null hypothesis.
The 95 % confidence interval around the point estimate (xbar = 25) is
25 +- 2.447(.75593) = 25+-1.849
95%CI is 23.15 < μ < 26.849. Notice the null value (24) is contained in the CI.
1c
H0 = 20 and H1 ≠ 20
With α = .05 reject H0 if t6 < -2.447 or > 2.447
Since a t of – 4.410 is less than – 2.447 (p = .005) we reject the null hypothesis.
The 95 % confidence interval around the point estimate (xbar = 15) is
15 +- 2.447(1.13) = 15 +- 2.76
95%CI is 12.24 < μ < 17.76. Notice the null hypothesis value (20) is not contained in the 95%CI.
1d. H0 ≤ 10and H1 > 10
With α = .01 reject H0 if t6 > 3.14
Since a t of 1.323 is less than 3.14 (p = .234/2 = .11) we do not reject the null hypothesis.
1e
H0 = 23 and H1 ≠ 23
With α = .01 reject H0 if t7 < -3.44 or > 3.44
Since a t of 2.82 is greater than – 3.44 and less than 3.44 (p = .025) we do not reject the null hypothesis.
The 95% confidence interval around the point estimate (xbar = 25) is
25 +- 2.36(.707) = 25 +- 1.66
95%CI is 23.34 < μ < 26.66 Notice the null hypothesis value (20) is not contained in the 95%CI. If we had used α = .05 we would have rejected.
2. The makers of ALPO would like to know whether eating ALPO dog food increases dogs’ energy. To test this they assessed the energy of a group of dogs, fed them ALPO and then assessed their energy a second time. Energy was measured using a 5 point scale with 1 = very low energy and 5 = very high energy. Below are the results of the study.
Lassie / 1 / 2 / -1 / 1
Brian / 2 / 4 / -2 / 4
Remo / 2 / 4 / -2 / 4
Rasta / 3 / 5 / -2 / 4
Moco / 1 / 1 / 0 / 0
Sum
/ 9 / 16 / -7 / 13Mean / 1.8 / 3.2 / -1.4
a) Perform the appropriate test to determine whether the energy gain is statistically significant (one-tailed test with α = .05).
Hand Calculation:
Vard = N∑d2 – (∑d)2
N(N-1)
Vard = 5(13) –(-7)2
5(4)
= 65 - 49
20
= .80
sd = .80^.5 = .89
s.e.d = .89/5^.5 = .4
t = / s.e.d
t = -1.4 / .4 = -3.5
t (critical one-tailed; alpha = .05, df = 4) = 2.132. Calculated t = 3.50. Statistically significant p < .05.
With SPSS
SPSS Paired Samples T-Test
b) Based on the above study the management team at ALPO concludes that ALPO improves dogs’ energy levels over that of other dog foods. What is wrong with their conclusion?
The sample statistics suggest that the average canine energy level on a scale of 1-5 increased by 1.4 increments by the end of the study. This suggests about a 25% increase in energy level over time. While this experiment did appear to show an effect over time, there was no comparison made among brands of dog food other than ALPO. Therefore, one cannot draw a conclusion that allows one to COMPARE ALPO with other brands. The finding simply shows that dogs energy level increases over time when fed.