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lana [24]
3 years ago
7

A tropical storm has been moving at 15 miles per hour for the past two days. Bess recorded that the storm moved 135 miles yester

day and 75 miles today. For how many hours has bess been keeping trrack of the storm?
Mathematics
1 answer:
Allushta [10]3 years ago
8 0

Answer:

Bess has been keeping track of the storm for 14 hours.

Step-by-step explanation:

The tropical storm has been moving at 15 miles per hour.

Bess recorded that the storm moved 135 miles yesterday and 75 miles today. Total distance covered by storm is

135+75=210

The relationship between distance speed and time is defined as

speed=\frac{distance}{time}

15=\frac{210}{time}

time=\frac{210}{15}

time=14

Therefore, Bess has been keeping track of the storm for 14 hours.

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Answer:

The null hypothesis: \mathbf{H_o: p=0.27}

The alternative hypothesis: \mathbf{H_1: p \neq 0.27}

Test statistics : z = −2.30

P-value:  = 0.02144

Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.

Conclusion: We accept the alternative hypothesis and  conclude that under the same​ circumstances the proportion of offspring peas will be yellow is not equal to 0.27

Step-by-step explanation:

From the given information:

Let's state the null and the alternative hypothesis;

Since The claim is that 27%  of the offspring peas will be yellow.

The null hypothesis state that the proportion of offspring peas will be yellow is equal to 0.27.

i.e

\mathbf{H_o: p=0.27}

The alternative hypothesis  state that the proportion of offspring peas will be yellow is not equal to 0.27

\mathbf{H_1: p \neq 0.27}

<u>The test statistics:</u>

we are given 437 green peas and 129 yellow apples;

Hence;

\hat p = \dfrac{x}{n}

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n = total number of the sample size

\hat p = \dfrac{129}{437+129}

\hat p = \dfrac{129}{566}

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Now; the test statistics can be computed as :

z = \dfrac { \hat p -p }{\sqrt {\dfrac{p(1-p)}{n}  } }

z = \dfrac {0.2279 -0.27 }{\sqrt {\dfrac{0.27(1-0.27)}{566}  } }

z = \dfrac {-0.043 }{\sqrt {\dfrac{0.27(0.73)}{566}  } }

z = \dfrac {-0.043 }{\sqrt {\dfrac{0.1971}{566}  } }

z = \dfrac {-0.043 }{\sqrt {3.48233216*10^{-4} } }

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P-value = P(Z < z)

P-value = P(Z< -2.30)

By using the ​ P-value method and the normal distribution as an approximation to the binomial distribution.

from the table of standard normal distribution

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find the probability value as 0.010724 by the intersection of the row and column values gives the area to the left of

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Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.

Conclusion: We accept the alternative hypothesis and  conclude that under the same​ circumstances the proportion of offspring peas will be yellow is not equal to 0.27

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