Question

a.Construct the confidence interval for the population mea μ.

c=0.98 x=16.3​, σ=5.0, and n=95

A 98​% confidence interval for μ is ( ___ , ___ ) ​(Round to one decimal place as​ needed.

b. Find the critical value tc for the confidence level c=0.98 and sample size n=13.

tc= _____ ​(Round to the nearest thousandth as​ needed.)

c. Find the margin of error for the given values of​ c, s, and n.c=0.80​, s=4​, n=28

The margin of error is _____ ​(Round to one decimal place as​ needed.)

Answer 1

a). Since we do not know the variance, we need to use the t-distribution and the first step is to calculate the standard error,

standard error (SE) = s / sqrt of the population size = 5 / sqrt (95) = 0.5130
also the degree of freedom (DF) is 95 - 1 = 94

we now reference the t distribution for a DF of 94 for 98% confidence interval is 2.367,
we can now calculate the lower and upper ranges of the 98% confidence interval
Lower Limit = 16.3 - 2.367 * 0.5130 = 15.09, or 15.1
Upper Limit = 16.3 + 2.367 * 0.5130 = 17.51, or 17.5

2.367

b). you want a 98% confidence interval with a sample size of 13.

df = n−1 = 12, and the area in one tail is (100%−98%)÷2 = 1% or 0.01, so you need t(12,0.01).

Hence, t_c = 2.681

c). ME = z * [ $\sigma$ /(n)^1/2]

= 1.333 x [5/(95)^1/2]

= 1.333 x [5/9.75]

= 1.333 x 0.5130 = 0.6838, 68.4%