Answer:
90% confidence the Population mean number of texts per day
(42.2561 ,47.1439)
Step-by-step explanation:
Step(i):-
Given sample size 'n' = 147
mean of the sample size x⁻ = 44.7
standard deviation of the sample 'S' = 17.9
90% confidence the Population mean number of texts per day
[tex](x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,(x^{-} + t_{\alpha } \frac{S}{\sqrt{n} })[/tex]
Step(ii):-
Degrees of freedom
ν=n-1=147-1=146
t₀.₁₀ = 1.6554
[tex](x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,(x^{-} + t_{\alpha } \frac{S}{\sqrt{n} })[/tex]
[tex](44.7 - 1.6554 \frac{17.9}{\sqrt{147} } ,(44.7 + 1.6554 \frac{17.9}{\sqrt{147} })[/tex]
(44.7 - 2.4439 ,44.7 + 2.4439 )
(42.2561 ,47.1439)
Conclusion:-
90% confidence the Population mean number of texts per day
(42.2561 ,47.1439)
