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Given a sample mean difference of 23, a standard error of 4.2, and a critical t value of 2.262, what are the values for the conf

erastova [34]

Answer: (13.4996, 32.5004)

Step-by-step explanation:

The confidence interval for mean difference is given by :-

$(\overline{x}_1-\overline{x}_2)\pm t^*\times SE$ ,

where $(\overline{x}_1-\overline{x}_2)$ = sample mean difference .

$t^*$ = critical t value (two-tailed)

SE= Standard error.

Given : $(\overline{x}_1-\overline{x}_2)=23$

$t^*=2.262$

SE= 4.2

Now, the confidence interval for mean difference will be :-

$23\pm (2.262)\times (4.2)\\\\= 23\pm9.5004\\\\=(23-9.5004,\ 23+9.5004)\\\\=(13.4996,\ 32.5004)$

Hence, the required confidence interval : (13.4996, 32.5004)

8 0

3 years ago

3 Given the function f (x) = 3.84 – 5x2 + 2x – 3, evaluate for f(

Artemon [7]

Step-by-step explanation:

f(x) = 3x⁴ - 5x² + 2x - 3

When x = -1,

f(-1) = 3(-1)⁴ - 5(-1)² + 2(-1) - 3 = 3 - 5 - 2 - 3 = -7.

3 0

3 years ago

How do you do this. Please give a full answer and explain why.

OverLord2011 [107]

Law of Exponent

$\displaystyle \large{ {a}^{ - n} = \frac{1}{ {a}^{n} } }$

Compare the terms.

$\displaystyle \large{ {a}^{ - n} = {( - 2)}^{ - 3} }$

Therefore, a = -2 and n = 3. From the law of exponent above, we receive:

$\displaystyle \large{ {( - 2)}^{ - 3} = \frac{1}{ {( - 2)}^{ 3} } }$

Exponent Def. (For cubic)

$\displaystyle \large{ {a}^{3} = a \times a \times a }$

Factor (-2)^3 out.

$\displaystyle \large{ {( - 2)}^{ - 3} = \frac{1}{( - 2) \times ( - 2) \times ( - 2)}}$

(-2) • (-2) = 4 | Negative × Negative = Positive.

$\displaystyle \large{ {( - 2)}^{ - 3} = \frac{1}{4 \times ( - 2)}}$

4 • (-2) = -8 | Negative Multiply Positive = Negative.

$\displaystyle \large{ {( - 2)}^{ - 3} = \frac{1}{ - 8}}$

If either denominator or numerator is in negative, it is the best to write in the middle or between numerator and denominators.

Hence,

$\displaystyle \large \boxed{ {( - 2)}^{ - 3} = - \frac{1}{ 8}}$

The answer is - 1 / 8

3 0

3 years ago

-4t=15=-1 what is t

Alexandra [31]

Answer:

-4

Step-by-step explanation:

-4t+15=-1

Subtract 15 on both sides:

-4t=-16

Divide -4 on both sides:

t=-4

3 0

3 years ago

Given that it was less than 80º on a given day, what is the probability that it also rained that day? 0.3 0.35 0.65 0.7

eimsori [14]

The probability that it also rained that day is to be considered as the 0.30 and the same is to be considered.

<h3>What is probability?</h3>

The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.

The probability that the temperature is lower than 80°F and it rained can be measured by determining the number at the intersection of a temperature that less than 80°F and rain.

So, This number is 0.30.

Hence, we can say that it was less than 80°F on a given day, the probability that it also rained that day is 0.30.

To learn more about the probability from the given link:

brainly.com/question/18638636

The above question is incomplete.

The conditional relative frequency table was generated using data that compared the outside temperature each day to whether it rained that day. A 4-column table with 3 rows titled weather. The first column has no label with entries 80 degrees F, less than 80 degrees F, total. The second column is labeled rain with entries 0.35, 0.3, nearly equal to 0.33. The third column is labeled no rain with entries 0.65, 0.7, nearly equal to 0.67. The fourth column is labeled total with entries 1.0, 1.0, 1.0. Given that it was less than 80 degrees F on a given day, what is the probability that it also rained that day?

#SPJ4

7 0

2 years ago