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Statistics help please!?

An investigation was conducted to determine the average amount of time children (ages 6-8) watching television each week. A sample of 25 children gave an average of 18 hours per week, with a 5:00 gap type. To determine a confidence interval 90% in one might expect to find the true mean (μ) for all children (ages 6-8).

ANSWER: 90% confidence interval results for the "truth": = [16.3, 19.7] Why? Small sample, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION x-bar = Half of the sample n = 18 s = standard sample number of samples deviation5 25 DF Degrees of freedom = 24 Level90 digits1 significant confidence "Look-up table 't-critical value'1 0.7 Look-Up Table t critical values for confidence intervals and prediction. central area on both sides = 90% with df = 24. Look-up Another method is to use Microsoft Excel function: TINV (Chance, degrees_freedom) Returns the inverse of the Student t distribution 90% confidence interval results for the "real" bar x + / – ('t critical value) * s / sqrt (n) = 18 + / – 1.7 * 5/SQRT (25) = [16.3, 19.7]

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