The sample was selected in a way that was expected to result in a sample representative of americans in this age group. (a) of those surveyed, 1123 reported owning a cell phone. use this information to construct a 90% confidence interval estimate of the proportion of all americans age 8 to 18 who own a cell phone. (round your answers to three decimal places.)

I am 90% confident that all Americans age 11.273 to 14.727 own a cell phone

Explanation:

Age data: 8,9,10,11,12,13,14,15,16,17,18

From the data above,

Mean is 13 and Standard deviation is 3.162

Number of sample (n) = 11

Degree of freedom = n - 1 = 11 - 1 = 10

Significance level = 1 - confidence level = 1 - 0.9 = 0.1

Use the t-distribution table to obtain the t-value at half the significance level (0.05) and 10 degree of freedom, t-value is 1.812

Confidence interval is given by [mean + (t-value × standard deviation) ÷ √number of sample] and [mean - (t-value × standard deviation) ÷ √number of sample

Confidence interval = 13 + (1.812×3.162)÷√11 = 13+(5.730÷3.317)= 13 + 1.727= 14.727

Confidence interval = 13 - (1.812×3.162)÷√11 = 13 - (5.730÷3.317) = 13 - 1.727 = 11.273

Confidence interval is 11.273 to 14.727. I am 90% confident that Americans age 11.273 to 14.727 own a cell phone

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< 3

send them to therapy or somthing like that