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 find the minimum sample size for the student union's confidence interval, we can use the formula:
n = (Z * sqrt(p * (1 - p))) / E
Where:
n = sample size
Z = z-score corresponding to the desired confidence level (in this case, 90% confidence level)
p = estimated proportion (0.5 can be a conservative estimate when the proportion is unknown)
E = margin of error
In this case, the margin of error should be at most 0.1, so E = 0.1. The Z-score for a 90% confidence level is approximately 1.645.
Plugging in these values into the formula, we get:
n = (1.645 * sqrt(0.5 * (1 - 0.5))) / 0.1
Simplifying this equation, we find that the minimum sample size required is 270.125, but since you can't have a fraction of a person, the student union should interview at least 271 full-time students.
Now, let's move on to the second question about finding a confidence interval for the difference between the gestational periods of group A and group B.
To find a confidence interval for the difference between the mean gestational periods of group A and group B, we can use the formula for the confidence interval of the difference of two means:
CI = (X̄1 - X̄2) ± (t * SE)
Where:
CI = Confidence interval
X̄1 = Mean of group A
X̄2 = Mean of group B
t = t-score corresponding to the desired confidence level and degrees of freedom
SE = Standard error of the difference
First, let's calculate the means for group A and group B.
Mean for Group A:
X̄1 = (40 + 41 + 38 + 40 + 40 + 39 + 39 + 41 + 41 + 40 + 40 + 40 + 39 + 38) / 14 ≈ 39.36
Mean for Group B:
X̄2 = (38 + 30 + 32 + 42 + 30 + 38 + 35 + 32 + 38 + 39 + 29 + 29 + 32) / 13 ≈ 34.23
Next, let's calculate the standard deviation for both groups.
Standard Deviation for Group A:
S1 = sqrt(((40-39.36)^2 + (41-39.36)^2 + (38-39.36)^2 + (40-39.36)^2 + (40-39.36)^2 + (39-39.36)^2 + (39-39.36)^2 + (41-39.36)^2 + (41-39.36)^2 + (40-39.36)^2 + (40-39.36)^2 + (40-39.36)^2 + (39-39.36)^2 + (38-39.36)^2) / (14-1)) ≈ 1.14
Standard Deviation for Group B:
S2 = sqrt(((38-34.23)^2 + (30-34.23)^2 + (32-34.23)^2 + (42-34.23)^2 + (30-34.23)^2 + (38-34.23)^2 + (35-34.23)^2 + (32-34.23)^2 + (38-34.23)^2 + (39-34.23)^2 + (29-34.23)^2 + (29-34.23)^2 + (32-34.23)^2) / (13-1)) ≈ 4.16
Since the question assumes equal variances, we can pool the variances:
Sp^2 = ((14-1) * (1.14^2) + (13-1) * (4.16^2)) / (14 + 13 - 2) ≈ 10.35
Using the pooled variance, we can calculate the standard error of the difference:
SE = sqrt(Sp^2 * (1/14 + 1/13)) ≈ 1.75
Now, let's determine the critical t-score for a 95% confidence level with degrees of freedom (df) equal to the sum of the sample sizes minus two:
df = 14 + 13 - 2 = 25
Using a t-table or calculator, the critical t-score for a 95% confidence level and 25 degrees of freedom is approximately 2.064.
Finally, plug in all the values into the confidence interval formula:
CI = (39.36 - 34.23) ± (2.064 * 1.75)
The lower bound of the confidence interval is calculated as (39.36 - 34.23) - (2.064 * 1.75) ≈ 1.83
The upper bound of the confidence interval is calculated as (39.36 - 34.23) + (2.064 * 1.75) ≈ 9.97
Therefore, the 95% confidence interval for the difference between the gestational periods of group A and group B is approximately 1.83 to 9.97 (group A mean minus group B mean).
Assumptions made:
1. The samples are representative of their respective populations.
2. The data is independent and random.
3. The population variances of both groups are equal.