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Nuetrik [128]
3 years ago
15

|29. Are the figures similar? Explain.10

Mathematics
2 answers:
Bond [772]3 years ago
6 0
They are not similer because they are facing different ways and they are different sizes
Alexeev081 [22]3 years ago
3 0
Right... they are not the similar, I ca know that because one is obliviously smaller than the other one
You might be interested in
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).

7 0
3 years ago
For a fair coin, suppose you toss the coin 100 times.
Natasha_Volkova [10]

Using the normal distribution, it is found that there is a 0.0005 = 0.05% probability of getting more than 66 heads.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
  • The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with \mu = np, \sigma = \sqrt{np(1-p)}.

For the binomial distribution, the parameters are given as follows:

n = 100, p = 0.5.

Hence the mean and the standard deviation of the approximation are given as follows:

  • \mu = np = 100(0.5) = 50.
  • \sigma = \sqrt{np(1-p)} = \sqrt{100(0.5)(0.5)} = 5

Using continuity correction, the probability of getting more than 66 heads is P(X > 66 + 0.5) = P(X > 66.5), which is <u>one subtracted by the p-value of Z when X = 66.5</u>.

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

Z = \frac{66.5 - 50}{5}

Z = 3.3

Z = 3.3 has a p-value of 0.9995.

1 - 0.9995 = 0.0005.

0.0005 = 0.05%

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

6 0
1 year ago
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scoray [572]
Hi hello how are you
8 0
2 years ago
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Please help!!! The scatterplot below shows the number of bags of popcorn sold when x movie tickets are sold.
ElenaW [278]
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The upper right corner is that point. You can simply read the graph.
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3 years ago
Find The Perimeter please asap
nevsk [136]

Answer:

44

Step-by-step explanation:

8 0
3 years ago
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