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lapo4ka [179]
3 years ago
12

Direct mail advertisers send solicitations​ ("junk mail") to thousands of potential customers in the hope that some will buy the

​ company's product. The response rate is usually quite low. Suppose a company wants to test the response to a new flyer and sends it to 1030 people randomly selected from their mailing list of over​ 200,000 people. They get orders from 103 of the recipients.
Required:
Create a 95% confidence interval for the percentage of people the company contacts who may buy something. (Show your work. Step by step)
Mathematics
1 answer:
harina [27]3 years ago
5 0

Answer:

The interval is (.0817, .118).

Step-by-step explanation:

Let's make this interval! The easy way is to go to STAT->TEST->A, 1-prop z-int, but I will show the long way.

<u>1. Conditions</u>

First we must check the conditions.

<em>Randomization condition: </em>The sample is given to be random, so we can assume independence.

<em>Success/failure condition: </em>Both np>10 and nq>10 must be met.

  • np = 1030(103/1030) = 103 ✓
  • nq = 1030(927/1030) = 927 ✓

<em>10% condition: </em>The sample cannot be greater than 10% of the population.

  • 10n < N, 10(1030) < 200000 ✓

All conditions are met, so we can use <u>a 1-proportion z-interval</u> to represent the data.

<u></u>

<u>2. Mechanics</u>

<u>Find </u>\hat{p}<u>, the sample proportion.</u>

103/1030 = .1

\hat{q}, then, the probability of not p, is 1 - .1 = .9

<u>Find z* (z star), the critical z-value.</u>

You can do this on your calculator using 2nd->VARS->3 (or you can memorize it - it's 1.96. Winky face.) This is called the invNorm function, that takes the left side area under the normal curve. But what do we plug in to the invNorm function?

invNorm(.975) = 1.96

<u>Plug values into the equation.</u>

\hat{p} ± z*\sqrt{\frac{\hat{p}\hat{q}}{n} }

.1 ± 1.96\sqrt{\frac{.1*.9}{1030} }

.1 ± 1.96(.00935)

.1 - .0183 = .0817

.1 + .0183 = .118

(.0817, .118)

<u>3. Conclusion</u>

Based on this sample, we are 95% confident that the proportion of people the company contacts who may buy something is between .0817 and .118. That isn't very likely, but maybe they'll make a little profit!

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A self proclaimed psychic was tested for ESP. The psychic was presented with 200 cards face down and asked to determine if the c
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Answer:

The 95% confidence interval would be given (0.172;0.288).  

We are confident at 95% that the true probability that the psychic correctly identifies the symbol on the card in a random trial is between (0.172;0.288).  

We can conclude that her random guessing would have got her correct 20% of the time. Since the confidence interval contains 0.2, she has no psychic powers, rather she just guessed it like normal people. Any person could have got it correct this many times.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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Description in words of the parameter p

p represent the probability that the psychic correctly identifies the symbol on the card in a random trial

\hat p represent the estimated probability that the psychic correctly identifies the symbol on the card in a random trial

n=200 is the sample size required  

z_{\alpha/2} represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Numerical estimate for p

In order to estimate a proportion we use this formula:

\hat p =\frac{X}{n} where X represent the cases successful

\hat p=\frac{46}{200}=0.23 represent the estimated probability that the psychic correctly identifies the symbol on the card in a random trial

Confidence interval

The confidence interval for a proportion is given by this formula:

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96  

And replacing into the confidence interval formula we got:  

0.23 - 1.96 \sqrt{\frac{0.23(1-0.23)}{200}}=0.172  

0.23 + 1.96 \sqrt{\frac{0.23(1-0.23)}{200}}=0.288  

And the 95% confidence interval would be given (0.172;0.288).

We are confident at 95% that the true probability that the psychic correctly identifies the symbol on the card in a random trial is between (0.172;0.288).  

We can conclude that her random guessing would have got her correct 20% of the time. Since the confidence interval contains 0.2, she has no psychic powers, rather she just guessed it like normal people. Any person could have got it correct this many times.

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