salary study conducted. mean for men = $42,000. Std deviation = $9,000. mean for women = $39,000. Std dev for women = 8,000. If 100 men surveyed, at a CI of 95% what is the the most number of women surveyed if there was no Statistical difference in the mean for men and women?
use the formula
42-39)/(8/root100)
work out z value
then 1 -95% = 5% 0.05 on table which is 1.6445
To determine the most number of women surveyed if there was no statistical difference in the mean for men and women, we need to use the concept of the confidence interval. The confidence interval is a range around the estimated sample mean within which the true population mean is likely to be found.
In this case, we have the following information:
For men:
- Sample mean = $42,000
- Standard deviation = $9,000
- Number of men surveyed = 100
For women:
- Sample mean = $39,000
- Standard deviation = $8,000
We want to find the most number of women surveyed at a 95% confidence interval (CI), assuming no statistical difference in the mean for men and women.
Here's how we can calculate it step by step:
1. Calculate the standard error (SE) for each group:
- For men: SE_men = standard deviation_men / sqrt(number of men surveyed)
- For women: SE_women = standard deviation_women / sqrt(number of women surveyed)
2. Calculate the margin of error (MOE) for each group at a 95% confidence interval:
- For men: MOE_men = 1.96 * SE_men (multiplied by 1.96 because we want 95% CI)
- For women: MOE_women = 1.96 * SE_women
3. Set up the equation using the means and margins of error:
- mean_men - mean_women = MOE_men + MOE_women
4. Solve the equation for the number of women surveyed:
- number of women surveyed = (mean_men - mean_women - MOE_men) / MOE_women
Let's calculate it now:
SE_men = $9,000 / sqrt(100) = $900
SE_women = $8,000 / sqrt(number of women surveyed)
MOE_men = 1.96 * $900 = $1,764
MOE_women = 1.96 * ($8,000 / sqrt(number of women surveyed))
mean_men - mean_women = $42,000 - $39,000 = $3,000
Setting up the equation:
$3,000 = $1,764 + 1.96 * ($8,000 / sqrt(number of women surveyed))
Now, we can solve this equation to find the number of women surveyed:
$3,000 - $1,764 = 1.96 * ($8,000 / sqrt(number of women surveyed))
$1,236 = 1.96 * ($8,000 / sqrt(number of women surveyed))
sqrt(number of women surveyed) = ($8,000 * 1.96) / $1,236
sqrt(number of women surveyed) ≈ 12.71
Now, squaring both sides:
number of women surveyed ≈ 12.71^2 ≈ 161.23
Therefore, the maximum number of women surveyed for which there would be no statistical difference in the mean salaries for men and women at a 95% confidence interval is approximately 161.