Nicole wants to find out which sport is the fifth grade students' favorite to play in gym class. She decides to do a survey.

Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):

Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, $\mu_X$ represent the population mean for Brand B and let $\mu_Y$ represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

P.Q. = $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

where, $Xbar$ = Sample mean for Brand B data = 36.9

$Ybar$ = Sample mean for Brand A data = 32.9

$n_1$ = Sample size for Brand B data = 13

$n_2$ = Sample size for Brand A data = 9

$s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }$ = 3.013

Here, $s^{2}_X$ and $s^{2} _Y$ are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < $t_2_0$ < 2.528) = 0.98

P(-2.528 < $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.528) = 0.98

P(-2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(Xbar -Ybar) -(\mu_X-\mu_Y)$ < 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

P( (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(\mu_X-\mu_Y)$ < (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ , (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ]

[ $(36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ , $(36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

4 0

2 years ago

If Q = {all perfect squares less than 30} and p={ all odd numbers from 1 to 10}

Soloha48 [4]

When Q = {all perfect squares less than 30} and p={ all odd numbers from 1 to 10) Q ∩ P = { 1, 9}

<h3>How to calculate the value?</h3>

Set theory simply means the branch of mathematical logic that deals with sets, that can be described as collections of objects.

It should be noted that perfect squares are the numbers that can be divided to give same number.

Q = {all perfect squares less than 30}. This will be 1, 4, 9, 16, 25

P ={ all odd numbers from 1 to 10}. This will be 1, 3, 5, 7, and 9.

In this case, the common numbers to set of P and Q are 1 and 9.

Therefore, the numbers are 1 and 9 since they're common to both sides.

Learn more about numbers on:

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If Q = {all perfect squares less than 30} and p={ all odd numbers from 1 to 10}. Find Q ∩ P.

3 0

1 year ago

Are there any ethical responsibilities that

Naya [18.7K]

Answer:

no...................

8 0

2 years ago

X2 + 5x - 50 = 0<br><img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%2B%205x%20-%2050%20%3D%200" id="TexFormula1" ti

love history [14]

Answer:

x = -10

x = 5

Step-by-step explanation:

${x}^{2} + 5x - 50 = 0 \\ (x + 10)(x - 5) = 0 \\ \\ x = - 10 \\ x = 5$

8 0

2 years ago

( − 35 − 42 − 63 ) ∶ ( + 7 )

olga nikolaevna [1]

(-35-42-63)/7= (-140)/7=-20

7 0

2 years ago