A student union wants to obtain a 90% confidence interval for the proportion of “full time” students who hold a part-time job for 20 hours or more per week. For this purpose, they want to interview a sample of full-time students. They want a margin of error to be at most 0.1, i.e., the length of the confidence interval should not exceed 0.2. What is the minimum number of full-time students they should interview?
• A researcher wants to determine the difference between the gestational period (period of pregnancy) for two groups of women, namely, those who had normal pregnancies (group A) and those who suffered from a certain complication during pregnancy known as preeclampsia (group B). The following data shows the gestational periods (in weeks) for a sample of 14 women from group A and of 13 women from group B.
Gestational periods for sample from group A: 40, 41, 38, 40, 40, 39, 39, 41, 41, 40, 40, 40, 39, 38
Gestational periods for sample from group B: 38, 30, 32, 42, 30, 38, 35, 32, 38, 39, 29, 29, 32
Find a 95% confidence interval for the difference between the gestational periods of group A and group B, i.e., group A mean minus group B mean. Assume that the variances of the two groups are equal. State any other assumptions you make
To determine the minimum sample size for the student union's study, we can use the formula for sample size calculation for estimating a proportion, given a desired confidence level and margin of error. The formula is:
n = (Z^2 * p * (1-p)) / E^2
Where:
- n represents the sample size
- Z is the z-value corresponding to the desired confidence level (in this case, 90% confidence level)
- p represents the estimated proportion
- E is the desired margin of error
In this case, the desired margin of error is 0.1, and the confidence level is 90%. Since we don't have an estimate of the proportion, we can use the worst-case scenario estimate, which is 0.5 (maximum variability). The z-value for a 90% confidence level is approximately 1.645.
Plugging in the values:
n = (1.645^2 * 0.5 * (1-0.5)) / 0.1^2
n = 0.426 / 0.01
n = 42.6
Since you cannot have a fraction of a student, we need to round up, giving us a minimum sample size of 43 full-time students.
For the researcher with the gestational periods, to find the confidence interval for the difference between the mean gestational periods of group A and group B, we will use the formula for calculating the confidence interval for the difference between two means:
CI = (X̄₁ - X̄₂) ± Z * √[ (s₁^2/n₁) + (s₂^2/n₂) ]
Where:
- CI represents the confidence interval
- X̄₁ and X̄₂ are the means of group A and group B, respectively
- Z is the z-value corresponding to the desired confidence level (in this case, 95% confidence level)
- s₁ and s₂ are the standard deviations of group A and group B, respectively
- n₁ and n₂ are the sample sizes of group A and group B, respectively
First, we calculate the means, standard deviations, and sample sizes for the two groups:
Group A:
Mean (X̄₁) = (40+41+38+40+40+39+39+41+41+40+40+40+39+38) / 14
= 39.857
Standard Deviation (s₁) = √[Σ(Xᵢ - X̄)² / (n₁ - 1)]
= √[Σ(40-39.857)² + (41-39.857)² + ... + (38-39.857)² / (14-1)]
= √[0.030 / 13]
≈ 0.161
Sample Size (n₁) = 14
Group B:
Mean (X̄₂) = (38+30+32+42+30+38+35+32+38+39+29+29+32) / 13
≈ 34.308
Standard Deviation (s₂) = √[Σ(Xᵢ - X̄)² / (n₂ - 1)]
= √[Σ(38-34.308)² + (30-34.308)² + ... + (32-34.308)² / (13-1)]
= √[35.256 / 12]
≈ 2.346
Sample Size (n₂) = 13
Next, we need to find the z-value for a 95% confidence level. It is approximately 1.96.
Plugging all the values into the formula:
CI = (39.857 - 34.308) ± 1.96 * √[(0.161²/14) + (2.346²/13)]
CI = 5.549 ± 1.96 * 0.105
CI = 5.549 ± 0.206
CI ≈ (5.343, 5.755)
Therefore, the 95% confidence interval for the difference between the gestational periods of group A and group B is approximately (5.343, 5.755).