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ludmilkaskok [199]
2 years ago
11

Acapulco, Mexico and Hyderabad, India both lie at 17° north latitude, and lie very nearly halfway around the world from each oth

er in an east-west direction. The radius of Earth at a latitude of 17° is about 3790 miles. Suppose that you could fly from Acapulco directly west to Hyderabad or fly directly north to Hyderabad. Which way would be shorter, and by how much? Use 3960 miles for Earth’s radius. (Hint: To fly directly north, you would go from 17° north latitude to 90° north latitude, and then back down to 17° north latitude.)
Mathematics
1 answer:
Nataly_w [17]2 years ago
6 0
In solving this problem we can consider Earth to be a sphere. When we have a circle, then we can use this formula to find arc length:
L=2 \pi R \frac{C}{360}
Where:
L= arc length (in this problem it is disance we need to travel)
R = radius of circle (in this problem it is equal to a radius of Earth)
C = angle we need to pass

We are told that two cities lie halfway around the world. If we fly to west this means angle is 180°.
This gives an arc length of:
L=2* \pi *3960* \frac{180}{360}  \\  \\ L=12440.71 miles

If we want to fly to north we need to go to 90° northern latidtude and then back to 17° latitude. This means angle is:
C=2*(90-17)=2*73°=146°
This gives an arc length of:
L=2* \pi *3960* \frac{146}{360} \\ \\ L=10090.8 miles

We can see that flying north is shorter. It is shorter by:
12440.71 miles - 10090.8 miles = 2349.91 miles
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A company wants to find out if the average response time to a request differs across its two servers. Say µ1 is the true mean/ex
lorasvet [3.4K]

Answer:

a) The null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0

Test statistic t=0.88

The P-value is obtained from a t-table, taking into acount the degrees of freedom (419) and the type of test (two-tailed).

b)  A P-value close to 1 means that a sample result have a high probability to be obtained due to chance, given that the null hypothesis is true. It means that there is little evidence in favor of the alternative hypothesis.

c) The 95% confidence interval for the difference in the two servers population expectations is (-0.372, 0.972).

d) The consequences of the confidence interval containing 0 means that the hypothesis that there is no difference between the response time (d=0) is not a unprobable value for the true difference.

This relate to the previous conclusion as there is not enough evidence to support that there is significant difference between the response time, as the hypothesis that there is no difference is not an unusual value for the true difference.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that there is significant difference in the time response for the two servers.

Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0

The significance level is 0.05.

The sample 1, of size n1=196 has a mean of 12.5 and a standard deviation of 3.

The sample 2, of size n2=225 has a mean of 12.2 and a standard deviation of 4.

The difference between sample means is Md=0.3.

M_d=M_1-M_2=12.5-12.2=0.3

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{3^2}{196}+\dfrac{4^2}{225}}\\\\\\s_{M_d}=\sqrt{0.046+0.071}=\sqrt{0.117}=0.342

Then, we can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.3-0}{0.342}=\dfrac{0.3}{0.342}=0.88

The degrees of freedom for this test are:

df=n_1+n_2-1=196+225-2=419

This test is a two-tailed test, with 419 degrees of freedom and t=0.88, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>0.88)=0.381

As the P-value (0.381) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that there is significant difference in the time response for the two servers.

<u>Confidence interval </u>

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=196 has a mean of 12.5 and a standard deviation of 3.

The sample 2, of size n2=225 has a mean of 12.2 and a standard deviation of 4.

The difference between sample means is Md=0.3.

The estimated standard error of the difference is s_Md=0.342.

The critical t-value for a 95% confidence interval and 419 degrees of freedom is t=1.966.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_{M_d}=1.966 \cdot 0.342=0.672

Then, the lower and upper bounds of the confidence interval are:

LL=M_d-t \cdot s_{M_d} = 0.3-0.672=-0.372\\\\UL=M_d+t \cdot s_{M_d} = 0.3+0.672=0.972

The 95% confidence interval for the difference in the two servers population expectations is (-0.372, 0.972).

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3 years ago
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