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1 February, 19:15

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.)

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  1. 1 February, 20:26
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    I am 90% confident that all Americans age 11.273 to 14.727 own a cell phone

    Explanation:

    Age dа ta: 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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