Question

A researcher is interested in finding a 90% confidence interval for the mean number of times per

Answer 1

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

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

Step(ii):-

Degrees of freedom

       ν=n-1=147-1=146

t₀.₁₀ =  1.6554

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

$(44.7 - 1.6554 \frac{17.9}{\sqrt{147} } ,(44.7 + 1.6554 \frac{17.9}{\sqrt{147} })$

(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)