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Assignment06_OROUR - 7.16 a) xbar=(/30 2249/30=74.967...

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7.16 a) Give a point estimate for the mean charitable commitment of tax-exempt organizations: x-bar =  (charitable commitment) / 30 2249/30 = 74.967 Because n is large (n = 30), by the Central Limit Theorem x-bar is a point estimator for the  mean charitable commitment of tax-exempt organizations. b) Construct a 98% confidence interval for the mean charitable commitment: x-bar  ±  z α /2  ( σ x-bar ) 74.967  ±  2.32 (18.02199/ 30) 74.967  ±  7.634 = (67.333, 82.601) Using the large sample confidence interval equation for  µ , I find that the mean charitable  commitment lies in the interval of 67.333 to 82.601 with 98% confidence. This is given the  following the conditions: that the random sample was selected from the target population and  that the sample size n is large (n   30). Given that n is large (n = 30), by the Central Limit  Theorem it is guaranteed that the sampling distribution of x-bar is approximately normal,  therefore we can have confidence that  µ  lies in the interval at least 98% of the time. c) What assumption must hold for the method of estimation used in part b to be appropriate: That the random sample was selected from the target population and that the sample size n is  large (n   30). d) Why is the confidence interval of part b a better estimator of the mean charitable commitment  than the point estimator of part a? This is because in reviewing the sample chartable commitment it is noticed that not all  charitable commitments lie within the interval range. Some are as low as 15%. Given that it is 
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