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ivolga24 [154]
3 years ago
13

According to the University of Nevada Center for Logistics Management, 6% of all merchandise sold in the United States gets retu

rned. A Houston department store sampled 80 items sold in January and found that 8 of the items were returned. (a) Construct a point estimate of the proportion of items returned for the population ofsales transactions at the Houston store. If required, round your answer to three decimal places.
Mathematics
1 answer:
KatRina [158]3 years ago
8 0

Answer:  0.1

Step-by-step explanation:

Given :  A Houston department store sampled 80 items sold in January and found that 8 of the items were returned.

In other words, sample size : <em>n</em>=1040

Number of items returned : <em>x</em>= 8

Let <em>p</em> be the proportion of items returned for the population of sales transactions at the Houston store.

As per sample , the sample proportion of items returned for the population of sales transactions at the Houston store is :

\hat{p}=\dfrac{x}{n}=\dfrac{8}{80}=0.1

As we know that , <em>the sample proportion is the best estimate of the population proportion.</em>

Therefore,a point estimate of the proportion of items returned for the population of sales transactions at the Houston store is 0.1.

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Answer:

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

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For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Proportion of 0.6

This means that p = 0.6

Sample of 46

This means that n = 46

Mean and standard deviation:

\mu = p = 0.6

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6*0.4}{46}} = 0.0722

Probability of obtaining a sample proportion less than 0.5.

p-value of Z when X = 0.5. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0.5 - 0.6}{0.0722}

Z = -1.38

Z = -1.38 has a p-value of 0.0838

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

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