A
A pharmaceutical company wants to know if a new vitamin supplement improves memory.
Study A: Researchers surveyed 2,000 people aged 60+. They found that those who took vitamin supplements daily performed 15% better on memory tests than those who did not.
Study B: Researchers recruited 200 volunteers. They randomly assigned 100 to take the vitamin and 100 to take a sugar pill (placebo) for 6 months. Both groups took a memory test at the end.
Which study is an experiment, and which is observational?
In Study A, can we conclude the vitamins caused better memory? Why or why not?
Study A is Observational; Study B is an Experiment.
No. Correlation does not imply causation. There could be confounding variables (e.g., people who take vitamins might also exercise more or eat healthier).
You want to estimate the average price of a cappuccino in the city. You sample 6 cafes:
{4.50, 5.00, 4.00, 5.50, 6.00, 4.50}
Describe the steps to create a single bootstrap sample.
Draw 6 numbers from the original list, with replacement. Calculate the mean of those 6 numbers
A marketing team wants to see if "Age Group" affects "Preferred Social Media". Total sample size = 500. TikTok | Facebook | Total |
| Under 30 | 150 | 50 | 200 |
| Over 30 | 100 | 200 | 300 |
Total | 250 | 250 | 500 |
Calculate the Expected Count for the (Under 30, Facebook) cell.
E = 10.
df= {rows}-1){cols}-1) = (2-1)(2-1) = 1
An IT help desk tracks the number of "Critical Severity" tickets received per hour. The maximum they ever get is 3
Tickets (x) | 0 | 1 | 2 | 3 |
| P(x) | 0.40 | 0.30 | 0.20 | ? |
Find the value of P(x=3)
1 - (0.4 + 0.3 + 0.2) = 0.10.
you want to estimate the average price of a cappuccino in the city. You sample 6 cafes:
{4.50, 5.00, 4.00, 5.50, 6.00, 4.50}
You generate 1,000 bootstrap means. The standard deviation of these means (Standard Error) is 0.28. The original sample mean is 4.92. Construct a 95% Confidence Interval.
Sample Mean 4.92
4.92 +- 2(0.28)
We compare the battery life (in hours) of 3 different mobile phone brands (Brand A, Brand B, Brand C).
SSBetween = 50 (df = 2)
SError = 200 (df = 20)
State the Null Hypothesis.
Calculate MSBetween and MSError
Calculate the F-statistic.
H_0: mu_A = mu_B = mu_C (All mean battery lives are equal).
MS_{Between} = 50 / 2 = 25.
MS_{Error} = 200 / 20 = 10.
F = 25 / 10 = 2.5.
The height of adult men is normally distributed with population mean = 175 cm and population standard deviation = 10 cm.
If you take a sample of n=25 men, what is the mean and standard deviation (Standard Error) of the sampling distribution of the sample mean?
How does the Standard Error change if you increase n to 100?
SE= sigma/ sqrt(n)
SE=10/sqrt(25)
SE=2
NEW SE= 10/ sqrt(100)
NEW SE=1
A factory produces light bulbs. A quality control check of 400 bulbs finds that 20 are defective.
Calculate the sample proportion of defective bulbs.
Construct a 95% Confidence Interval for the true defect rate.
phat= 20/400 = 0.05
SE=sqrt((0.05*0.95)/400)
CI= 0.05+- 1.96*(SE)
(0.029, 0.071) or2.9% to 7.1%.
6 participants join a weight-loss program. Weights are measured Before and After.
Differences = {After} - {Before}
{-2, -1, -3, 0, -4, -2\}
Mean difference = -2.0.
SD of difference = 1.41
State the Null and Alternative hypotheses to test if the program reduces weight.
Calculate the t-statistic.
Degrees of freedom (df) for this test?
H_0: mu_d = 0 (No change).
H_a: mu_d < 0 (Weight decreased; numbers are negative).
t = -2.0/1.41 / sqrt{6} =approx -3.48.
df = n - 1 = 6 - 1 = 5.