An advertising media analyst wants to estimate the mean weekly amount of time consumers’ spend watching...
An advertising media analyst wants to estimate the mean weekly
amount of time consumers’ spend watching television daily. Based on
previous studies, the standard deviation is assumed to be 20
minutes. The media analyst wasn’t to estimate, with 99% confidence,
the mean weekly amount of time to within +/- 5 minutes.
a) What sample sizes is needed
b) If 95% confidence is desired, how many consumers need to be
selected?
Homework Answers
DATE:29-07-2020
ANSWER:--------
![a). for 99.7. Confidence interval, 2= 2.576. Hence Sample size needed. 2 n = (20 - (2.526x24] [SD-02 20 mg E = +5. (10.304)](http://img.homeworklib.com/questions/a62731e0-0daf-11eb-ba68-3d6b93eec06b.png?x-oss-process=image/resize,w_560)
![6). for 95% CI, 2=1.96 Hence, sample size needed. n- 2013 (1-96x20] = (7.84² = 61.46 ~ 62. Hence decrealeng ci level decrease](http://img.homeworklib.com/questions/a76cf180-0daf-11eb-a317-fd6a614b6558.png?x-oss-process=image/resize,w_560)
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