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Given the following information about a hypothesis test of the difference between two means based on independent random samples,

HACTEHA [7]

Answer:

$S^2_p =\frac{(13-1)(5)^2 +(10 -1)(3)^2}{13 +10 -2}=18.143$

And the deviation would be just the square root of the variance:

$S_p=4.259$

Then the statistic is given by:

$t=\frac{(12 -9)-(0)}{4.259\sqrt{\frac{1}{13}+\frac{1}{10}}}=1.674$

And the correct option would be:

t = 1.674

Step-by-step explanation:

Data given:

$n_1 =13$ represent the sample size for group 1

$n_2 =10$ represent the sample size for group 2

$\bar X_1 =12$ represent the sample mean for the group 1

$\bar X_2 =9$ represent the sample mean for the group 2

$s_1=5$ represent the sample standard deviation for group 1

$s_2=3$ represent the sample standard deviation for group 2

We are assuming two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

The pooled variance is given by:

$S^2_p =\frac{(13-1)(5)^2 +(10 -1)(3)^2}{13 +10 -2}=18.143$

And the deviation would be just the square root of the variance:

$S_p=4.259$

Then the statistic is given by:

$t=\frac{(12 -9)-(0)}{4.259\sqrt{\frac{1}{13}+\frac{1}{10}}}=1.674$

And the correct option would be:

t = 1.674

4 0

4 years ago

What is a quartic function with only two real zeros atx x = 7 and x = 13

Strike441 [17]

Answer:

b

Step-by-step explanation:

B. Y= (x - 10) ^ 4 - 81 OR y = x^4 - 20x +91

4 0

3 years ago

How many solutions are in the equation below?

Elis [28]

Answer:

A. 1

Step-by-step explanation:

12x + 6 = 5x

12x - 5x = - 6

7x = - 6

x = - 6 / 7

Only one solution.

8 0

3 years ago

Read 2 more answers

Use the Alternating Series Approximation Theorem to find the sum of the series sigma^infinity_n = 1 (-1)^n - 1/n! with less than

DanielleElmas [232]

Answer:

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198 -0.0000248$

For the 7th term we have 3 decimals of approximation but our value is 0.000198 higher than the error required, so we can use the 8th term and we have that $|-0.0000248|= 0.0000248$ and with this we have 4 decimals of approximation so if we add the first 8 terms we have a good approximation for the series with an error bound lower than 0.0001.

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198-0.0000248 =0.632118$

Step-by-step explanation:

Assuming the following series:

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!}$

We want to approximate the value for the series with less than 0.0001 of error.

First we need to ensure that the series converges. If we have a series $\sum a_n$ where $a_n = (-1)^n b_n$ [/tex] or $a_n =(-1)^{n-1} b_n$ where $b_n \geq 0$ for all n if we satisfy the two conditions given:

1) $lim_{n \to \infty} b_n =0$

2) { $b_n$ } is a decreasing sequence

Then $\sum a_n$ is convergent. For this case we have that:

$lim_{n \to \infty} \frac{1}{n!} =0$

And $\frac{1}{n!}$ because $\frac{1}{n!} =\frac{1}{n (n-1)!}$ and $\frac{1}{n(n-1)!} < \frac{1}{(n-1)!}$

So then we satisfy both conditions and then the series converges. Now in order to find the approximation with the error required we can write the first terms for the series like this:

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198 -0.0000248$

For the 7th term we have 3 decimals of approximation but our value is 0.000198 higher than the error required, so we can use the 8th term and we have that $|-0.0000248|= 0.0000248$ and with this we have 4 decimals of approximation so if we add the first 8 terms we have a good approximation for the series with an error bound lower than 0.0001.

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198-0.0000248 =0.632118$

6 0

4 years ago

Add write your answer in simplest form 5 3/4+2 11/12

olya-2409 [2.1K]

Answer: 26/3 or 8 2/3

Step-by-step explanation:

In the addition or subtraction of fraction, first, you have to convert all of them into improper fraction if they are mixed; then, you need to find the LCD (Least Common Denominator).

<u>Converting fraction:</u>

5 3/4=23/4

2 11/12=35/12

<u>Finding LCD:</u>

4 and 12 is 12

<u>Converting fraction into common denominator:</u>

23/4=69/12

35/12=35/12

<u>Solve:</u>

5 3/4 + 2 11/12

=69/12 + 35/12

=104/12

4 0

4 years ago