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aksik [14]
2 years ago
15

How many observations are there for each case in a t test for dependent samples?

Mathematics
1 answer:
Oxana [17]2 years ago
5 0

The number of observations for each case in a t test for dependent samples is two is the correct answer.

In this question,

The dependent t-test also called the paired t-test or paired-samples t-test compares the means of two related groups to determine whether there is a statistically significant difference between these means. Each sample must be randomly selected from a normal population and each member of the first sample must be paired with a member of the second sample.

A dependent samples t-test uses two raw scores from each person to calculate difference scores and test for an average difference score that is equal to zero.

The groups contain either the same set of subjects or different subjects that the analysts have paired meaningfully. In dependent samples, subjects in one group do provide information about subjects in other groups.

Hence we c an conclude that the number of observations for each case in a t test for dependent samples is two is the correct answer.

Learn more about dependent t-test here

brainly.com/question/15870238

#SPJ4

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Answer: 12

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QUESTION 2<br>Simplify the following<br>a5×a7​
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a^{12}

Step-by-step explanation:

<em><u>Identity Used </u></em>:   a^x \times a^y = a^{x+ y}

   

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In order to ensure efficient usage of a server, it is necessary to estimate the mean number
juin [17]

Answer:

a. [36.19;39.21]

b. Reject the null hypothesis. The population mean of users that are connected at the same time is greater than 35.

Step-by-step explanation:

Hello!

Your study variable is,

X: "number of users of one server at a time"

The objective is to estimate the mean, for this, a sample of n=100 times was taken and the standard deviation S= 9.2 and the sample mean is X[bar]= 37.7 were calculated.

You need to study the population mean, for this you need your variable to have at least normal distribution. Since you don't have information about its distribution, but the sample is big enough (n≥30) you can apply the Central Limit Theorem and approximate the distribution of the sample mean X[bar] to normal:

X[bar]≈N(μ;σ²/n)

a. With this approximation, you can construct the 90% Confidence Interval using the approximate Z

[X[bar] ± Z_{1-\alpha /2} * S/√n]

Z_{1-\alpha /2} = Z_{0.95} = 1.64

[37.7± 1.64* 9.2/√100]

[36.19;39.21]

b. You need to test if the population mean is greater than 35 with a level of significance of 1%.

The hypothesis is:

H₀: μ ≤ 35

H₁: μ > 35

α: 0.01

This is a one-tailed test so you have only one critical level (right tail):

Z_{1\alpha } = Z_{0.99} = 2.33

This means that if the value of the calculated statistic is equal or greater than 2.33 you will reject the null Hypothesis.

If the value is less than 2.33 you will support the null hypothesis.

The statistic is:

Z=<u> X[bar] - μ </u>= <u> 37.7 - 35 </u> = 2.93

       S/√n           9.2/10

The value 2.93 > 2.33, so you reject the null hypothesis. This means that the population mean of users that are connected at the same time is greater than 35.

<u><em>Note: </em></u><em>To make the decision using the interval calculated on a), the hypothesis should have been two-tailed and the confidence and significance levels complementary.</em>

I hope it helps!

7 0
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Okay, so first of all, if the trapezoid was translated right 4,
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</span>(6,-6)(8,-6)(9,-5)(5,-5)
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