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2 years ago

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A researcher is interested in determining if the more than two thirds of students would support making the Tuesday before Thanks

giving a holiday. The researcher asks 1,000 random selected students if they would support making the Tuesday before Thanksgiving a holiday. Seven hundred students said that they would support the extra holiday. What is the null and alternative hypothesis?

1 answer:

Yakvenalex [24]2 years ago

3 0

Answer:

The null hypothesis is, {H₀}: p = 0.66, H₀: p = 0.66 .

The alternative hypothesis is, {Hₐ}: p > 0.66, Hₐ: p > 0.66

Step-by-step explanation:

Let p represent the proportion of students who are supportive of making the day before Thanksgiving a holiday.

Since two populations are considered, the alternative hypothesis for the difference of the means of the two populations has to be stated.

The proportion of two-third students are supportive of making the day before Thanksgiving a holiday is p = 0.66 obtained as shown below:

The null hypothesis is,

{H₀}:p = 0.66, H₀: p = 0.66

The alternative hypothesis is,

{Hₐ:p > 0.66, Ha: p > 0.66

The null hypothesis is, {H₀}: p = 0.66, H₀: p=0.66 .

The alternative hypothesis is, {Hₐ}: p > 0.66, Hₐ: p > 0.66

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Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

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In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that $\mu = 8.3, \sigma = 3.3$ .

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We are working with a sample mean of 37 jets. So we have that:

$s = \frac{3.3}{\sqrt{37}} = 0.5425$

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$X = \frac{320}{37} = 8.65$

This probability is the pvalue of Z when $X = 8.65$ . So

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

$Z = \frac{8.65 - 8.3}{0.5425}$

$Z = 0.65$

$Z = 0.65$ has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

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Total time of 275 minutes for 37 jets, so

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This probability is subtracted by the pvalue of Z when $X = 7.43$

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

$Z = \frac{7.43 - 8.3}{0.5425}$

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$X = \frac{320}{37} = 8.65$

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$X = \frac{275}{37} = 7.43$

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