Enter Title

A

100

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).

200

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.

300

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

B

100

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

200

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)

300

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%.

C

100

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.

• state the null hypothesis and alternate hypothesis

E   = 10.

df= {rows}-1){cols}-1) = (2-1)(2-1) = 1 

200

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.

300

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.