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Gnom [1K]
2 years ago
6

Assume that all conditions are met. The mean of the differences was 1. 33 and the standard deviation of those differences was 2.

90. What is the test statistic for this procedure?.
Mathematics
1 answer:
Maurinko [17]2 years ago
8 0

The test statistic for this procedure is using two sample T-test is 0.458.

Definition of Two Sample T-test

A variation of the Student's t-test used to determine if two sample means are substantially different is known as the two sample T-test (also known as Welch's t-test, Welch's adjusted T, or unequal variances t-test). The test's degrees of freedom have been changed, which generally boosts the test's power for samples with unequal variance.

T-Test Statistic for The Procedure

It is given that mean of the differences = 1.33

And, the standard deviation of the differences, standard error = 2.90

The test static formula is given as,

t = (estimate - hypothized value) / standard error

Here,  t is the t- test statistic, and estimate is the mean of the differences.

Since it is given that all conditions are met, the hypothized value can be taken as 0

Substituting substituting estimate = 1.33 and standard error = 2.90 in the t-test statistic formula, we get,

t = (1.33-0)/2.90

or t = 0.458

learn more about t-test here:

brainly.com/question/14128303

#SPJ4

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Answer:

6x+15

Step-by-step explanation:

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3 years ago
According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe
Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1 - p)}{n}}, as long as np \geq 10 and n(1 - p) \geq 10.

In this problem:

  • 22% of couples meet online, hence p = 0.22.
  • A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

\mu = p = 0.22

s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 0.25</u>, hence:

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

By the Central Limit Theorem:

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

Z = \frac{0.25 - 0.22}{0.0338}

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the <u>p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15</u>, hence:

X = 0.2:

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

Z = \frac{0.2 - 0.22}{0.0338}

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

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

Z = \frac{0.15 - 0.22}{0.0338}

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

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Step-by-step explanation:

15 = 3x

x = 15 : 3

x = 5

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