Question

The height (in inches) of men at UH is assumed to have a normal distribution with a standard deviation of 3.6 inches. The height (in inches) of women at UH is also assumed to have a normal distribution with a standard deviation of 2.9 inches. A random sample of 49 men and 38 women yielded respective means of 68.3 inches and 64.6 inches. Find the 90% confidence interval for the difference in the heights of men at UH and women at UH

Answer 1

Answer: The required 90% confidence interval would be (2.55,4.85).

Step-by-step explanation:

Since we have given that

Standard deviation of men = 3.6 inches

Standard deviation of women = 2.9 inches

Average of men = 68.3 inches

Average of women = 64.6 inches

Number of men = 49

Number of women = 38

α = 0.10

$\dfrac{\alpha }{2}=\dfrac{0.10}{2}=0.05\\\\So,\\\\z_{0.05}=1.645$

So, the confidence interval would be

$(\bar{x_1}}-\bar{x_2}})\pm z\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}\\\\=(68.3-64.6)\pm 1.645\sqrt{\dfrac{3.6^2}{49}+\dfrac{2.9^2}{38}}\\\\=3.7\pm 1.645\times 0.69\\\\=(3.7-1.15,3.7+1.15)\\\\=(2.55,4.85)$

Hence, the required 90% confidence interval would be (2.55,4.85).