Question

weights of men: 90% confidence; n = 14, x=161.3 lb, s =12.6 lb

Answer 1

Answer:

The answer is below

Step-by-step explanation:

Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation . Assume that the population has a normal distribution. Round the confidence interval limits to the same number of  decimal places as the sample standard deviation.

Answer: Given that:

Sample size (n) = 14, mean ($\mu$) = 161.3, standard deviation ($\sigma$) = 12.6

Confidence(C)= 90% = 0.9

α = 1 - C = 1- 0.9 = 0.1

α/2 = 0.1 / 2 = 0.05

The z score of α/2 correspond to a z score of 0.45 (0.5 - 0.05). This gives:

$z_{\frac{\alpha}{2} }=1.645$

The margin of error (E) is given by the formula:

$E=z_{\frac{\alpha}{2} }\frac{\sigma}{\sqrt{n} } =1.645*\frac{12.6}{\sqrt{14} }=5.5$

The confidence interval = μ ± E = 161.3 ± 5.5 = (155.8, 166.8)

The confidence interval is between 155.8 lb and 166.8 lb. There is a 90% confidence that the mean is between 155.8 lb and 166.8 lb.