Question

A simple random sample of size nequals17 is drawn from a population that is normally distributed. The sample mean is found to be x overbar equals 56 and the sample standard deviation is found to be sequals10. Construct a 95​% confidence interval about the population mean. The lower bound is nothing. The upper bound is nothing. ​(Round to two decimal places as​ needed.)

Answer 1

Answer:

95% confidence intervals about the population mean is

(51.7656 , 60.2344)

Step-by-step explanation:

Step(i):-

Given random sample of size 'n' =17

Given mean of the sample 'x⁻' = 56

Given standard deviation of sample 's' = 10

95% confidence intervals about the population mean is determined by

$(x^{-} - t_{0.05} \frac{s}{\sqrt{n} } ,x^{-} + t_{0.05} \frac{s}{\sqrt{n} } )$

Degrees of freedom

ν = n-1 = 17-1 =16

t₀.₀₅ = 1.7459  (from t-table)

Step(ii):-

95% confidence intervals about the population mean is determined by

$(x^{-} - t_{0.05} \frac{s}{\sqrt{n} } ,x^{-} + t_{0.05} \frac{s}{\sqrt{n} } )$

$(56 - 1.7459 \frac{10}{\sqrt{17} } ,56 + 1.7459 \frac{10}{\sqrt{17} } )$

( 56 - 4.2344 , 56 + 4.2344)

(51.7656 , 60.2344)

Conclusion:-

95% confidence intervals about the population mean is

(51.7656 , 60.2344)