D

According to a newspaper survey, about 8% of the local residents order groceries online at least once a month. Suppose we took random samples of n= 60 from this population and computed the proportion that \widehat{p} of residents in each sample who order groceries online at least once a month. We can assume the newspaper's claim is true.
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Anonymous
0

Which of the following distributions is the best approximation of the sampling distribution of \widehat{p}?


1,


2.


3.



4.






Sangeetha Pulapaka
1

The correct option is option 2.


We are looking at which conditions does the sampling distributions of the sample proportions, looks, skewed to the left, skewed to the right, approximately normal, or uniform.


Our sample size n = 60, and p = 0.08

Expected successes: np = 60 (0.08) = 4.8

Expected failures ; n(1 - p) = 60 ( 1- 0.08) = 55.2n(1-p) = 15 ( 1 - 0.09) = 15(0.91) = 13.65


Since np\lt 10, and  n(1-p) \geq 10, the shape of the sampling proportion is not normal.

Now to find in which direction the shape of the sampling distribution is skewed.


We would expect only a right-skewed sampling distribution when the population proportion is so extreme relative to the sample size that there are fewer than 10 expected successes per sample. So, since our expected successes is less than 10, we have a right-skewed sampling distribution.


Qalaxia Master Bot
0

I found an answer from www.khanacademy.org

Normal conditions for sampling distributions of sample proportions ...


I thought that n > 30 is what it needs to be in order for any sample distribution of sample statistic to be normally distributed according the central limit theorem. So,


For more information, see Normal conditions for sampling distributions of sample proportions ...