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Shalnov [3]
3 years ago
10

After a series of rainy days, a meteorologist wants to compare the average daily rainfall over the past week between two cities.

The daily rainfall totals, in inches, for both cities are given below.
City A = 1.6, 3.8, 2.2, 0.4, 3.4, 2.6, 1.5
City B = 1.2, 4.5, 0.9, 3.6, 2.1, 1.1, 1.3

Which of the following statements are true regarding the data sets above?

I. The mean rainfall of city A is greater than the mean rainfall of city B.
II. The median rainfall of city B is half the median rainfall of city A.
III. The range of rainfall amounts for city A is greater than the range of rainfall amounts for city B.
IV. The median rainfall is less than the mean rainfall for both cities.

II and III

I and III

I and IV

II and IV
Mathematics
1 answer:
siniylev [52]3 years ago
6 0

Answer:

I and IV

Step-by-step explanation:

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Combine like terms

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Answer:

AC=6

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A sample of 26 offshore oil workers took part in a simulated escape exercise, and their escape time (unit: second) were observed
Anuta_ua [19.1K]

Answer:

t=\frac{370.69-360}{\frac{24.36}{\sqrt{26}}}=2.238  

The degrees of freedom are given by:

df = n-1= 26-1=25

And the p value would be:

p_v =P(t_{25}>2.238)=0.0172  

If we use a 5% of significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 360 second or 6 minutes. We need to be careful since if we use a significance level of 1% the result change

Step-by-step explanation:

Information given

\bar X=370.69 represent the sample mean

s=24.36 represent the sample standard deviation

n=26 sample size  

\mu_o =6*60 =360 s represent the value to verify

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to check if the true mean is at most 360 seconds, the system of hypothesis would be:  

Null hypothesis:\mu \leq 360  

Alternative hypothesis:\mu > 360  

The statistic for this case would be given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

We can replace in formula (1) the info given like this:  

t=\frac{370.69-360}{\frac{24.36}{\sqrt{26}}}=2.238  

The degrees of freedom are given by:

df = n-1= 26-1=25

And the p value would be:

p_v =P(t_{25}>2.238)=0.0172  

If we use a 5% of significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 360 second or 6 minutes. We need to be careful since if we use a significance level of 1% the result change

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Answer:

Im not sure

Step-by-step explanation:

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