Chapter 6
In 2014, the New York Yankees had a team batting average of µ = 245 (actually 0.245, but for simplicity, the decimal point is omitted here). Of course, the batting average varies from game to game, but assuming that the distribution of batting averages for 162 games is normal with a standard deviation of σ = 40 points.
If you randomly select one game from 2014, what is the probability that the team batting average was over 300?
If you randomly select one game from 2014, what is the probability that the team batting average was under 200?
z = X – μ
σ
300 – 245
40
55/40 =
= 1.375
p(z > 1.375) = 0.0838
z = X – μ
σ
200 – 245
40
−45/40 =
= –1.125
p(z < –1.125) = 0.1292
Conceptual Question
How does sample size influence standard error?
The larger the sample size, the smaller the standard error. The smaller the sample size, the larger the standard error. There is a negative relationship here. Basically, as one increases the other decreases.
A group of researchers is testing a drug, and these researchers come to the conclusion that there is an effect, but it actually does not have an effect. What kind of error is this?
Type I error
Conceptual Question
What is the important difference between a t-test and a z-test? (In terms of information we have access to)
What do we require if we are to calculate a z-test?
The important difference between a z-test and t-test is:
In order to calculate a z-test we require the population standard deviation. This is not a common value to have when we are conducting research and therefore the t-test is a more common statistic to calculate. So if the population std dev is missing from a question, then we are likely dealing with a t-test.
IQ test scores are standardized to produce a normal distribution with a mean of μ = 100 and a standard deviation of σ = 15. Find the proportion of the population in the following IQ category.
Very superior intelligence: IQ from 120 to 140
z = IQ – μ/σ =
140 – 100
15
40/15 = 2.6667
p(z > 2.6667) = 0.0038 FIND WHATS IN THIS TAIL DRAW GRAPH
The z-score for the upper bound of this interval is 2.6667, as found in the previous part. For the lower bound of this interval,
z = IQ – μ
σ
120 – 100
15
20/15 = 1.3333
p(z > 1.3333) = 0.0918
p(1.3333 < z < 2.6667) = 0.0918 – 0.0038 = 0.0874
What are the two conditions in which a DSM can be considered normal? Using the Central Limit Theorem?
1) Original Population is normal
2) Sample size is n=30 or greater
A group of researchers is testing a drug. These researchers come to the conclusion that there is not an effect, but it actually does have an effect. What kind of error is this?
Type II error
IQ test scores are standardized to produce a normal distribution with a mean of μ = 100 and a standard deviation of σ = 15. Find the proportion of the population in the following IQ category.
Average or normal intelligence: IQ from 90 to 109
For the upper bound of this interval,
z = IQ – μ
σ
IQ – μ/σ =
109 – 100
15
9/15 = 0.6000
For the lower bound of this interval,
z = IQ – μ
σ
90 – 100
15
−10/15 = –0.6667
p(z < –0.6667) = 0.2514 - located in the tail
p(z > 0.6000) = 0.2743 - located in the tail
Therefore, 1 = 100% of a proportion
p(–0.6667 < z < 0.6000) = 1 – 0.2514 – 0.2743 = 0.4743
OR
You could find the proportion between the mean and the x value and then add those values up.
Which would be 0.60 & 067
0.60=0.2257
0.67=0.2486
.2257+.2486=0.4743
CONCEPTUAL QUESTION
1) What is the null hypothesis?
2) What is the alternative hypothesis?
3) What does it mean if we reject the null hypothesis; is there an effect or not an effect?
1) Hypothesis of no effect, no change, no difference between means. Symbolized by H0
2) Opposite of H0. Hypothesis of an effect, change, or difference. Symbolized by H1.
3) There is an effect!
A sample is obtained from a population with a mean of µ = 40, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 44.5 with a variance of s² = 36.
Calculate Cohen's d and interpret the effect size.
d = 44.5-40/√36 = 0.75
The effect size in this case would be large!
A random sample of n=9 is selected from a normal population with a mean of μ = 40 and a standard deviation of σ = 6. After a treatment is administered to the individuals in the sample, the sample mean is found to be M = 37.
The researchers calculated this sample stat: z=-1.50
With an α = .01, would we reject or retain this hypothesis? Compare statistic to critical value and make a decision. Then interpret the results.
p > .01 zcrit=-2.33 & 2.33
We would retain the null hypothesis in this case, as the z the researchers calculated did not exceed the zcrit determined by the alpha level.
Therefore, it does not appear that the treatment administered has an effect.
Childhood participation in sports, cultural groups, and youth groups appears to be related to improved self-esteem for adolescents (McGee, Williams, Howden-Chapman, Martin, & Kawachi, 2006). In a representative study, a sample of n = 100 adolescents with a history of group participation is given a standardized self-esteem questionnaire. For the general population of adolescents, scores on this questionnaire form a normal distribution with a mean of μ = 50 and a standard deviation of σ = 15. The sample of group-participation adolescents had an average of M = 53.8.
Use a two-tailed test with α = .05.
Complete the five steps of hypothesis testing:
1) State the hypothesis
2) Set the decision rule/critical value
3) Compute sample statistic
4) Compare stat to crit value and make a decision
5) Interpret your results
1) H0 : μ = 50
H1 : μ ≠ 50
2) α = .05, two-tailed: p = .025 on both sides of distribution. Zcrit = 1.96 and -1.96
3) First we must calculate the standard error:
σM=σ/√n σM=15/√100 = 1.5
Next we must conduct the z-test for the sample mean we were provided:
Z= M-μ/σM Z= 53.8-50/1.5 = 2.53
4) Stat exceeded critical value. Therefore, we reject the null hypothesis.
5) Childhood participation in sports, cultural groups, and youth groups appears to have an effect on self-esteem of adolescents.
In a nondirectional hypothesis test, evaluate the effect of a treatment. A sample is obtained from a population with a mean of µ = 40, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 44.5 with a variance of s² = 36. The researchers have a sample that consists of n = 4 individuals. Conduct a 5-step hypothesis test. With an alpha level of α = .05?
We would use the t-stat here because we do not know the standard deviation of the population.
Step 1)
H0 u=40
H1 u≠40
Step 2) - Two-tailed due to nondirectional phrasing. df=3 Use the t-distribution table.
zcrit= 3.182 & -3.182
Step 3)
SM= √(s2/n) SM=√(36/4) = 3.00
t = M-μ/SM t =44.5-40/3.00 =1.50
Step 4)
Our calculated t-stat of 1.50 does not exceed our critical value of 3.182, so there is no significant effect. We retain the null hypothesis.
Step 5)
The treatment was not shown to have an effect.