Question

In a random sample of 28 ​people, the mean commute time to work was 31.7 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 80​% confidence interval for the population mean mu. What is the margin of error of mu​? Interpret the results.

Answer 1

Answer:

a)The 80% of confidence interval for the Population 'μ' is

(29.9121 ,33.4879)

b) Margin of error of mean '  'μ' is  =  1.7879

Step-by-step explanation:

step(i):-

Given random sample 'n' =28

The mean of the sample x⁻ =31.7 min

The standard deviation of the sample's' =7.2

Step(ii):-

80% of confidence intervals:

The 80% of confidence interval for the Population 'μ' is determined by

$(x^{-} - t_{\frac{\alpha }{2} } \frac{s}{\sqrt{n} } , x^{-} + t_{\frac{\alpha }{2} } \frac{s}{\sqrt{n} } )$

Degrees of freedom

γ = n-1 = 28-1 =27

$t_{\frac{\alpha }{2} } = t_{\frac{0.20}{2} } = t_{0.1} = 1.314$

The 80% of confidence interval for the Population 'μ' is determined by

$(x^{-} - t_{\frac{\alpha }{2} } \frac{s}{\sqrt{n} } , x^{-} + t_{\frac{\alpha }{2} } \frac{s}{\sqrt{n} } )$

$(31.7-1.314 X \frac{7.2}{\sqrt{28} } , 31.7+1.314 X\frac{7.2}{\sqrt{28} } )$

(31.7 -1.7879 , 31.7 +1.7879)

(29.9121 ,33.4879)

b)

Margin of error of mean is determined by

$M.E = {\frac{t_{\frac{\alpha }{2} } s}{\sqrt{n} } }$

$M.E = {\frac{1.314 X 7.2}{\sqrt{28} } }$

Margin of error =1.7879