Probability Practice
For their senior comp project, a psychology student conducted a poll to see if there were differences online dating use across three different age groups. The results can be found in the table below.
Age Group Use Don’t Use
18-29 45 12
30-49 8 50
50+ 7 49
Using the data from the table above, answer the following question
P(Online Dating)
𝑃 (Online Dating) = 60/171 = 0.3509 = ~35.09%
About _______% of cases received a score below 550 when a test distribution has M = 500 and s = 100.
69
A sample of scores has s = 10. In this sample, a score of X = 80 corresponds to z = -1.50. What is the sample mean?
95
For a population with M = 60 and s = 8, what is the X value corresponding to z = 1.55?
72.4
A statistical test led Dr. Jones to decide that his treatment had an effect when in fact it did not. What is this called?
Type I Error
For their senior comp project, a psychology student conducted a poll to see if there were differences in
online dating use across three different age groups. The results can be found in the table below.
Age Group Use Don’t Use
18-29 45 12
30-49 8 50
50+ 7 49
Using the data from the table above, answer the following question
P (Online Dating ∩ 18-29)
𝑃 (Online Dating ∩ 18 ―29) = 45/171 = .2632 = ~26.32%
What is the probability of randomly selecting a z-score less than z = 1.75 from a normal distribution
0.9599
A sample of scores has M = 50. In this sample, an X value of 70 corresponds to z = 2.00. What is the sample standard deviation?
10
A sample of n = 25 is selected from a population with μ = 73 and a treatment is administered to the sample. After treatment, the sample mean is M = 86 with a population standard deviation of 6. Based on this information, the size of the treatment effect, as measured by Cohen's d, is __________.
d = 2.17
Which of the statements below is FALSE...
a. Increasing the alpha level can increase power in a test.
b. Using a two-tailed test typically can increase power in comparison to a one-tailed test.
c. Increasing the sample size can increase power in a test.
d. Decreasing standard deviation can increase power in a test.
b. Using a two-tailed test typically can increase power in comparison to a one-tailed test.
For their senior comp project, a psychology student conducted a poll to see if there were differences in
online dating use across three different age groups. The results can be found in the table below.
Age Group Use Don’t Use
18-29 45 12
30-49 8 50
50+ 7 49
Using the data from the table above, answer the following question
P (Online Dating U 18-29)
𝑃 (Online Dating ∪ 18 ―29) = 72/171 = .4211 = ~42.11%
What proportion of a normal distribution corresponds to z-scores greater than +1.64?
0.0505
A sample has a standard deviation of s = 6. In this distribution, a score that is 18 points above the mean would correspond to a z-score of __________.
z = 3
A normal distribution has a mean of µ = 100 with σ = 20. If one score is randomly selected from this distribution, what is the probability that the score will have a value between X = 85 and X = 110?
0.4649
Which of the following is true?
a. A lower alpha level is associated with a higher |z| critical value.
b. A lower alpha level is associated with a lower |z| critical value.
c. A lower alpha level is associated with a higher |z| obtained value.
d. A lower alpha level is associated with a lower |z| obtained value.
a. A lower alpha level is associated with a higher |z| critical value.
For their senior comp project, a psychology student conducted a poll to see if there were differences in
online dating use across three different age groups. The results can be found in the table below.
Age Group Use Don’t Use
18-29 45 12
30-49 8 50
50+ 7 49
Using the data from the table above, answer the following question
P (Online Dating | 18-29)
𝑃 (Online Dating│18 ―29) = 0.2632 / 0.3333 = 0.7900 = ~79.00%
What proportion of the scores in a normal distribution have z-scores below z = -0.86?
0.1949
A sample with M = 85 and s = 12 is transformed into z-scores. After the transformation, the sample of z-scores will have a standard deviation of __________.
1.00
Alissa drives to work each morning and the trip takes an average of µ = 40 minutes. The distribution of driving times is approximately normal with a standard deviation of σ = 5 minutes. For a randomly selected morning, what is the probability that Alissa’s drive to work will take less than 30 minutes?
0.0228
What does a power analysis tell you?
How many participants are needed to detect a given effect with a given level of confidence.
For their senior comp project, a psychology student conducted a poll to see if there were differences in
online dating use across three different age groups. The results can be found in the table below.
Age Group Use Don’t Use
18-29 45 12
30-49 8 50
50+ 7 49
Using the data from the table above, answer the following question
P (50+ | Online Dating)
𝑃 (50 + │Online Dating) = 0.0409 / 0.3509 = .1167 = ~11.67%
The mean of population is 118.2 with a standard deviation of 12. What is the probability of selecting a sample of 36 individuals whose mean is greater than 123? (Hint: What is the p-value?)
0.0082
A distribution with M = 35 and s = 8 is being standardized so that the new mean and standard deviation will be M = 40 and s = 5. In the new, standardized distribution your score is X = 60. What was your score in the original distribution?
X = 67
Dr. Parrot is looking into the number of times a Kākāpō (a type of bird) makes a booming call to a potential mate. She has a sample size of 50. On average, her sample calls about 5 times with a standard deviation of 2.
Calculate and interpret a 95% Confidence Interval using this information.
CI: [4.45, 5.55]
If Dr. Parrot replicated this study with the same sample size, then 95% of the confidence intervals created have a chance of containing the true population mean.
What are you looking for when you find the probability of getting a positive test results assuming you have some type of disease?
Sensitivity