Question

From Barbie dolls to runway models, women in Western countries are exposed to unrealistically thin and arguably unhealthy body standards for their gender. A body mass index (BMI) between 18.5 and 24.9 is considered healthy. Using a 3D computer avatar, participants built what they considered the ideal body of an adult of their own gender. Below appears a summary of the results for a sample of 40 female heterosexual Caucasian undergraduate students from British universities that had been recruited by the researchers. Mean BMI was recorded to be 18.85 with standard deviation 1.75. Does the data provide evidence that young Caucasian women in British Universities, on average, aim for an unhealthy ideal body type (corresponding to a BMI less than 18.5)? Use α = 0.10.

Required:
a. Construct a 99% confidence interval for the mean BMI for young Caucasian women in British universities using R. Use at least 2 decimals in your answer.
b. Would a 99% CI (for the mean BMI) constructed using a smaller sample than the one used for part a) tend to be wider or narrower? Explain.

Answer 1

Answer:

(18.1361 ; 19.5639) ;

using a smaller sample will widen the margin

Step-by-step explanation:

Given :

Xbar = 18.85

Standard deviation, s = 1.75

Sample size, n = 40

Confidence interval is obtained using the relation :

Xbar ± Margin of error

Margin of Error = Zcritical * s/ √n

Zcritical at 99% = 2.58

Margin of error = 2.58 * (1.75/√40) = 0.7139

Lower boundary = 18.85 - 0.7139 = 18.1361

Upper boundary = 18.85 + 0.7139 = 19.5639

(18.1361 ; 19.5639)

B.)

The sample size, n being the denominator in the margin error formula increases the error value as it reduces or decreases in value Hence widens the interval. Conversely, increasing the sample size reduces the error margin value and ultimately narrows the interval.

Therefore, using a smaller sample will widen the margin