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VMariaS [17]
2 years ago
9

Suppose that a manager is interested in estimating the average amount of money customers spend in her store. After sampling 36 t

ransactions at random, she found that the average amount spent was $ 41.15 . She then computed a 90 % confidence interval to be between $ 38.01 and $ 44.29 .
Mathematics
1 answer:
musickatia [10]2 years ago
8 0

Answer:

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Step-by-step explanation:

Assuming this complete question: Which statement gives a valid interpretation of the interval?

The store manager is 90% confident that the average amount spent by the 36 sampled customers is between S38.01 and $44.29.

There is a 90% chance that the mean amount spent by all customers is between S38.01 and $44.29.

There is a 90% chance that a randomly selected customer will spend between S38.01 and $44.29.

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

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.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

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Answer:

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Step-by-step explanation:

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According to Eurostat, 55.2% of households in Italy have one child, the highest percentage in the Euro zone. A random sample of
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Answer:

0.04

Step-by-step explanation:

In this case we must use the following formula to be able to find the standard error, which would be the square root, of the probability for its complement divided by the population:

SE= \sqrt{\frac{p *(1-p)}{n} } \\

Now, knowing that p = 0.552 and n = 160 we replace and we are left with:

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Mopeds (small motorcycles with an engine capacity below 50 cm3) are very popular in Europe because of their mobility, ease of op
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Answer:

a) P(X

And we can find this probability using the normal standard table or excel and we got:

P(z

b) P(X>48)=P(\frac{X-\mu}{\sigma}>\frac{48-\mu}{\sigma})=P(Z>\frac{48-46.8}{1.75})=P(z>0.686)

And we can find this probability using the complement rule, the normal standard table or excel and we got:

P(z>0.686)=1- P(Z

c) P(46.8-1.5*1.75

And we can find this probability with this difference:

P(-1.5

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

P(-1.5

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the velocities of a population, and for this case we know the distribution for X is given by:

X \sim N(46.8,1.75)  

Where \mu=46.8 and \sigma=1.75

We are interested on this probability

P(X

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X

And we can find this probability using the normal standard table or excel and we got:

P(z

Part b

P(X>48)=P(\frac{X-\mu}{\sigma}>\frac{48-\mu}{\sigma})=P(Z>\frac{48-46.8}{1.75})=P(z>0.686)

And we can find this probability using the complement rule, the normal standard table or excel and we got:

P(z>0.686)=1- P(Z

Part c

P(46.8-1.5*1.75

And we can find this probability with this difference:

P(-1.5

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

P(-1.5

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So the answer would be "C"



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