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Anna71 [15]
3 years ago
6

Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety eq

uipment. Suppose an airline conducts a survey. Over Thanksgiving weekend, it surveys 6 flights from Boston to Salt Lake City to determine the number of babies on the flights. It determines the amount of safety equipment needed by the result of that study.
Select the things that were wrong with the way the survey was conducted. (Select all that apply.)

O The survey uses systematic sampling.

O Conducting the survey on a holiday weekend will not produce representative results.

O Choosing 6 flights represents a stratified sample.

O The survey was conducted using six similar flights.

O The survey would not be a true representation of the entire population of air travelers.
Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
4 0

Answer:

-Conducting the survey on a holiday weekend will not produce representative results

-The survey was conducted using six similar flights

-The survey would not be a true representation of the entire population of air travelers

Step-by-step explanation:

First of all, the sampling method that was used in this survey/study was a convenience sampling, were they just used the data that was readily available, which in this case were the 6 flights from Boston to Salt Lake City.

This sampling method is useful for pilot studies and for identifying tendencies, however, the obtained sample is not representative of the population, and because there is no criteria to organize the sample, (for example there was no fight with a different route taking into account) it is impossible to obtain statistical results that are precise.

And besides that, the fact that the survey was carried out over Thanksgiving weekend is also a factor that can directly affect the results, so it needs to be taken into account as a variable, which in this study was not.

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Answer:

The interval containing the middle-most 48% of sample means is between 218.59 to 221.41.

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 normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributied random variable X, with mean \mu and standard deviation \sigma, the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 220, \sigma = 13, n = 35, s = \frac{13}{\sqrt{35}} = 2.1974

Find the interval containing the middle-most 48% of sample means:

50 - 48/2 = 26th percentile to 50 + 48/2 = 74th percentile. So

74th percentile

value of X when Z has a pvalue of 0.74. So X when Z = 0.643.

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

By the Central Limit Theorem

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

0.643 = \frac{X - 220}{2.1974}

X - 220 = 0.643*2.1974

X = 221.41

26th percentile

Value of X when Z has a pvalue of 0.26. So X when Z = -0.643

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

-0.643 = \frac{X - 220}{2.1974}

X - 220 = -0.643*2.1974

X = 218.59

The interval containing the middle-most 48% of sample means is between 218.59 to 221.41.

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I found the complete question on Chegg, here is the full question:

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