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25. Flightstats, Inc., collects data onthe number of fights scheduled and the number of flightsflown at major airports throughout the United States. Flightstats data showed 56$\%$ offights scheduled at Newark, La Guardia, and Kennedy airports were flown during a three-day snowstorm (The Wall Street Journal, February $21,2006 )$ . All airlines say they alwaysoperate within set safety parameters-if conditions are too poor, they don't fly. The fol-lowing data show a sample of 400 scheduled flights during the snowstorm.Use the chi-square test of independence with a .05 level of significance to analyze the data.What is your conclusion? Do you have a preference for which airline you would choose tofly during similar snowstorm conditions? Explain.

There is sufficient evidence to support the claim that the variables are notindependent.

Intro Stats / AP Statistics

Chapter 11

Comparisons Involving Proportions and a Test of Independence

Descriptive Statistics

Confidence Intervals

The Chi-Square Distribution

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So to answer this question, here we are, given that we have a sample size of 401st of all, and we basically knew that 56% of the scheduled flights at Newark, LaGuardia and Kennedy airports were flown during this three day snowstorm. So it wants us to basically music high square test of independence with 0.5 level of significance to analyze the data. So first of all, a chi square test could be taken by finding the some of the observed frequency minus the expected frequency divided by the expected frequency. And the numerator here is going to be squared to which once we essentially find the valley here, So whatever value we find So that's gonna be a question mark here. We basically you can use this to First of all, also find the degrees of freedom So the degrees of freedom here can be taken by, um, calculating role minus one multiplied by calling minus one. So after this value, these two values have been found here. We basically look at AP table and from the P table. If we find that our value of P here is going to be lessens your 0.5 Um, sorry. It should be less than 0.5 rather, which is going to be the 5% of significant sample. So if it's less than that, then we can essentially just reject the null hypothesis. So in this particular case, there is going to be sufficient evidence for us to support the claim that these variables are not independent. So therefore, in this case, so in this case, um, we cannot, um, reject the null hypothesis.

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