Based on the information given for each of the following studies, decide whether to reject the null hypothesis. For each, give (a) the Z-score cutoff (or cutoffs) on the comparison distribution at which the null hypothesis should be rejected,(b) the Z score on the comparison distribution for the sample score, and (c) your conclusion. Assume that all populations are normally distributed.
Population
Study ì ó Sample Score p Tails of Test
A
100.0
10.0
80
.05
1 (low predicted)
B
100.0
20.0
80
.01
2
C
74.3
11.8
80
.01
2
D
16.9
1.2
80
.05
1 (low predicted)
E
88.1
12.7
80
.05
2
18. A researcher predicts that listening to music while solving math problems will make a particular brain area more active. To test this, a research participant has her brain scanned while listening to music and solving math problems,and the brain area of interest has a percentage signal change of 58. From many previous studies with this same math problems procedure (but not listening to music), it is known that the signal change in this brain area is normally distributed with a mean of 35 and a standard deviation of 10. (a) Using the .01 level, what should the researcher conclude? Solve this problem explicitly using all five steps of hypothesis testing, and illustrate your answer with a sketch showing the comparison distribution, the cutoff (or cutoffs),and the score of the sample on this distribution. (b) Then explain your answer to someone who has never had a course in statistics (but who is familiar with mean, standard deviation, and Z scores).
A. Study:
Population Mean (μ) = 100.0
Population Standard Deviation (σ) = 10.0
Sample Mean = 80
p = .05
Tails of Test = 1 (low predicted)
(a) Z-score cutoff:
Since it's a one-tailed test, we look up the Z-score in a Z-table for a .05 probability on the lower end, which is -1.645.
(b) Z-score for the sample score:
Z = (Sample Mean - Population Mean) / (Population Standard Deviation / sqrt(n))
Z = (80 - 100) / (10 / sqrt(80))
Z = -20 / (10 / sqrt(80))
Z = -20 / (10 / 8.944)
Z = -20 / 1.118
Z = -17.89
(c) Conclusion:
Since the Z-score of the sample (-17.89) is less than the Z-score cutoff (-1.645), we reject the null hypothesis. This means that there is a significant difference between the sample score and the population mean.
B. Study:
Population Mean(μ) = 100.0
Population Standard Deviation (σ) = 20.0
Sample Mean = 80
p = .01
Tails of Test = 2
(a) Z-score cutoff:
Since it's a two-tailed test, we look up the Z-score for a total probability of .01. We find the Z-scores at both ends: -2.576 and 2.576.
(b) Z-score for the sample score:
Z = (Sample Mean - Population Mean) / (Population Standard Deviation / sqrt(n))
Z = (80 - 100) / (20 / sqrt(80))
Z = -20 / (20 / 8.944)
Z = -20 / 2.236
Z = -8.945
(c) Conclusion:
Since the Z-score of the sample (-8.945) is less than the lower Z-score cutoff (-2.576), we reject the null hypothesis. This means that there is a significant difference between the sample score and the population mean.
C. Study:
Population Mean(μ) = 74.3
Population Standard Deviation (σ) = 11.8
Sample Mean = 80
p = .01
Tails of Test = 2
(a) Z-score cutoff:
Since it's a two-tailed test, we look up the Z-score for a total probability of .01. We find the Z-scores at both ends: -2.576 and 2.576.
(b) Z-score for the sample score:
Z = (Sample Mean - Population Mean) / (Population Standard Deviation / sqrt(n))
Z = (80 - 74.3) / (11.8 / sqrt(80))
Z = 5.7 / (11.8 / 8.944)
Z = 5.7 / 1.318
Z = 4.32
(c) Conclusion:
Since the Z-score of the sample (4.32) is greater than the upper Z-score cutoff (2.576), we reject the null hypothesis. This means that there is a significant difference between the sample score and the population mean.
D. Study:
Population Mean(μ) = 16.9
Population Standard Deviation (σ) = 1.2
Sample Mean = 80
p = .05
Tails of Test = 1 (low predicted)
(a) Z-score cutoff:
Since it's a one-tailed test, we look up the Z-score in a Z-table for a .05 probability on the lower end, which is -1.645.
(b) Z-score for the sample score: (not solvable with given data)
The sample mean given (80) is not realistic in this context, as it would far exceed the population mean (16.9). The provided data seems incorrect, so we cannot calculate the Z-score and subsequently cannot test the hypothesis.
E. Study:
Population Mean(μ) = 88.1
Population Standard Deviation (σ) = 12.7
Sample Mean = 80
p = .05
Tails of Test = 2
(a) Z-score cutoff:
Since it's a two-tailed test, we look up the Z-score for a total probability of .05. We find the Z-scores at both ends: -1.96 and 1.96.
(b) Z-score for the sample score:
Z = (Sample Mean - Population Mean) / (Population Standard Deviation / sqrt(n))
Z = (80 - 88.1) / (12.7 / sqrt(80))
Z = -8.1 / (12.7 / 8.944)
Z = -8.1 / 1.417
Z = -5.72
(c) Conclusion:
Since the Z-score of the sample (-5.72) is less than the lower Z-score cutoff (-1.96), we reject the null hypothesis. This means that there is a significant difference between the sample score and the population mean.
18.
a) Five steps of hypothesis testing:
1. Restate the research question as a null hypothesis and an alternative hypothesis.
Null Hypothesis: The brain area’s signal change will not be different when listening to music.
Alternate Hypothesis: The brain area’s signal change will be different when listening to music.
2. Determine the characteristics of the comparison distribution.
Mean (μ) = 35
Standard Deviation (σ) = 10
3. Determine the cutoff (or cutoffs) for rejecting the null hypothesis.
Using α = .01 (two-tailed test), we look up the Z-scores at both ends: -2.576 and 2.576.
4. Calculate the test statistic (Z-score) for the sample score.
Z = (Sample Mean - Population Mean) / (Population Standard Deviation / sqrt(n))
Z = (58 - 35) / (10 / sqrt(1))
Z = 23 / 10
Z = 2.3
5. Decide whether to reject the null hypothesis.
Since the Z-score (2.3) does not fall within the rejection zone (-2.576 to 2.576), we fail to reject the null hypothesis.
b) Explain to someone without statistical knowledge:
In this study, we tested whether listening to music while solving math problems had an effect on a participant’s brain activity. We compared the brain activity of the participant while listening to music to the average brain activity of people who’ve done the same task without music. We found that the brain activity increased, but not enough to conclude a significant change. Therefore, we can't confidently say that listening to music while solving math problems changes the activity in that brain area.
To answer the question, we need to go through the five steps of hypothesis testing.
Step 1: State the hypotheses.
The null hypothesis (H0) is the statement that the researcher wants to test and assume to be true initially. In this case, the null hypothesis would be that listening to music does not make the brain area more active. The alternative hypothesis (Ha) is the opposite of the null hypothesis and what the researcher hopes to demonstrate. Here, the alternative hypothesis would be that listening to music does make the brain area more active.
Step 2: Formulate an analysis plan.
In this step, we define the significance level and determine the test statistic to use. The significance level, denoted by alpha (α), is the probability of making a Type I error - rejecting the null hypothesis when it is actually true. In this case, the significance level is given as .01.
Step 3: Analyze sample data.
Given that we have a sample score of 58, we need to convert this score into a Z-score. To do this, we use the formula: Z = (x - μ) / σ, where x is the sample score, μ is the mean, and σ is the standard deviation. Using the provided mean of 35 and standard deviation of 10, we can calculate the Z-score as follows:
Z = (58 - 35) / 10 = 2.3
Step 4: Interpret the results.
Now, we need to compare the Z-score to the cutoff values on the comparison distribution. The cutoff values are determined based on the significance level and the tails of the test.
(a) For a two-tailed test with a significance level of .01, the cutoff values can be found using a Z-table, but assuming a normal distribution, we can approximate them as ±2.57 (since .01/2 = .005, and the closest Z-score for .005 is 2.57). Therefore, if the Z-score falls outside the range -2.57 to 2.57, we would reject the null hypothesis.
(b) The Z-score for the sample score in this study is 2.3. As it falls within the range -2.57 to 2.57, we would not reject the null hypothesis. Therefore, the conclusion would be that there is not enough evidence to support the claim that listening to music makes the brain area more active.
In simpler terms, the researcher's prediction was that listening to music while solving math problems would increase activity in a specific brain area. However, based on the data collected from the brain scans, the observed signal change of 58 did not deviate significantly from the expected signal change of 35. Therefore, we do not have enough evidence to say that listening to music makes the brain area more active.