A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained physical activity. What sample size should be obtained if she wishes the estimate to be within four percentage points with 90% confidence, assuming that
(a) she uses the estimates of 22.2% male and 18.8% female from a previous year?
(b) she does not use any prior estimates?
Well, well, well, a physical therapist who wants to investigate the difference between men and women in physical activity. This is quite an interesting topic, isn't it? Now let's dive into your questions, shall we?
a) If our dear therapist wants to use the estimates of 22.2% for males and 18.8% for females from a previous year, and if she wants to obtain an estimate within four percentage points with 90% confidence, we can use a formula known as the sample size formula for estimating a proportion.
First, we determine the critical value for a 90% confidence level. This critical value is approximately 1.645.
Next, we plug it into the formula:
Sample Size = (Critical Value)^2 * (Estimated Proportion * (1 - Estimated Proportion)) / (Margin of Error)^2
For males:
Sample Size = (1.645)^2 * (0.222 * (1 - 0.222)) / (0.04)^2 ≈ 318.76
For females:
Sample Size = (1.645)^2 * (0.188 * (1 - 0.188)) / (0.04)^2 ≈ 360.28
So, if our therapist wants to be within four percentage points with 90% confidence, she should aim for a sample size of about 319 for males and 360 for females.
b) Now, if our therapist wants to get the sample size without any prior estimates, we would need to use a different approach. In such a case, we use a conservative estimate of 50% for both males and females, as it yields the largest sample size.
Using the same formula, now with an estimated proportion of 0.5:
Sample Size = (1.645)^2 * (0.5 * (1 - 0.5)) / (0.04)^2 ≈ 422.53
So, if our therapist decides not to use any prior estimates and opts for a conservative assumption of 50% participation for both genders, she would need a sample size of about 423.
Hope that brings a smile to your face while dealing with sample sizes and proportions!
To determine the sample size for the study, we can use the formula for estimating sample size for proportions.
(a) With prior estimates:
The formula is given as:
n = (Z^2 * p * q) / E^2
Where:
n = sample size
Z =Z-value corresponding to the desired confidence level (90% confidence level has a Z-value of approximately 1.645)
p = proportion estimate (use the larger estimate, in this case, 22.2% male)
q = 1 - p (proportion estimate of the opposite group)
E = maximum error or margin of error in the estimate (4 percentage points or 0.04)
Plugging in the values:
n = (1.645^2 * 0.222 * (1 - 0.222)) / 0.04^2
Calculating this expression gives us the sample size required.
(b) Without prior estimates:
When there are no prior estimates available, we can assume a proportion of 0.5 (maximum variability) to obtain the largest required sample size.
Using the same formula as above:
n = (1.645^2 * 0.5 * 0.5) / 0.04^2
Calculate this expression for the sample size required.
Please specify if you would like the calculations for either case or any further assistance.
To determine the sample size needed for estimating the difference in proportions, we can use the formula:
𝑛 = 𝑧² * 𝑝(1−𝑝) / 𝑑²
Where:
- 𝑛 represents the sample size needed.
- 𝑧 represents the z-score corresponding to the desired confidence level. For a 90% confidence level, the z-score would be approximately 1.645.
- 𝑝 represents the estimated proportion from the previous year.
- 𝑑 represents the desired margin of error (in this case, 4 percentage points).
Let's calculate the sample size needed for each scenario:
(a) With prior estimates from a previous year:
For males, 𝑝 = 22.2% = 0.222
For females, 𝑝 = 18.8% = 0.188
Z-score for a 90% confidence level: 𝑧 ≈ 1.645
Margin of error: 𝑑 = 4 percentage points = 0.04
Sample size for males:
𝑛₁ = (1.645² * 0.222 * (1-0.222)) / (0.04²)
= 350.315 / 0.0016
≈ 218947.1875
Sample size for females:
𝑛₂ = (1.645² * 0.188 * (1-0.188)) / (0.04²)
= 350.315 / 0.0016
≈ 218947.1875
Since we need the sample size to be a whole number, we can round up to the nearest integer. Therefore, the sample size for both males and females would be approximately 218948.
(b) Without using prior estimates:
In this case, we assume that we don't have any previous estimates for the proportions. Hence, we can use 0.5 as the estimated proportion since it represents the maximum variability.
For both males and females, 𝑝 = 0.5
Z-score for a 90% confidence level: 𝑧 ≈ 1.645
Margin of error: 𝑑 = 4 percentage points = 0.04
Sample size for males and females:
𝑛 = (1.645² * 0.5 * (1-0.5)) / (0.04²)
= 350.315 / 0.0016
≈ 218947.1875
Rounding up to the nearest integer, the sample size would be approximately 218948.
Therefore, in both scenarios, the sample size needed to estimate the difference in proportions with a 90% confidence level and a margin of error of 4 percentage points is approximately 218948.