Complete Question:
Find the directional derivative of g(x,y) = [tex]x^2y^5[/tex]at the point (1, 3) in the direction toward the point (3, 1)
Answer:
Directional derivative at point (1,3), [tex]D_ug(1,3) = \frac{162}{\sqrt{8} }[/tex]
Step-by-step explanation:
Get [tex]g'_x[/tex] and [tex]g'_y[/tex] at the point (1, 3)
g(x,y) = [tex]x^2y^5[/tex]
[tex]g'_x = 2xy^5\\g'_x|(1,3)= 2*1*3^5\\g'_x|(1,3) = 486[/tex]
[tex]g'_y = 5x^2y^4\\g'_y|(1,3)= 5*1^2* 3^4\\g'_y|(1,3)= 405[/tex]
Let P = (1, 3) and Q = (3, 1)
Find the unit vector of PQ,
[tex]u = \frac{\bar{PQ}}{|\bar{PQ}|} \\\bar{PQ} = (3-1, 1-3) = (2, -2)\\{|\bar{PQ}| = \sqrt{2^2 + (-2)^2}\\[/tex]
[tex]|\bar{PQ}| = \sqrt{8}[/tex]
The unit vector is therefore:
[tex]u = \frac{(2, -2)}{\sqrt{8} } \\u_1 = \frac{2}{\sqrt{8} } \\u_2 = \frac{-2}{\sqrt{8} }[/tex]
The directional derivative of g is given by the equation:
[tex]D_ug(1,3) = g'_x(1,3)u_1 + g'_y(1,3)u_2\\D_ug(1,3) = (486*\frac{2}{\sqrt{8} } ) + (405*\frac{-2}{\sqrt{8} } )\\D_ug(1,3) = (\frac{972}{\sqrt{8} } ) + (\frac{-810}{\sqrt{8} } )\\D_ug(1,3) = \frac{162}{\sqrt{8} }[/tex]
Toby cuts a pizza into 6 equal slices. He eats half a slice. What fraction of the pizza has he eaten?
The pizza is cut into 6 slices so each slice would be 1/6 of the pizza.
He at 1/2 of a slice:
1/6 x 1/2 = 1/12 of the pizza
find the value of x if (1.1)^x=100
Answer:
x ≈ 48.3177
Step-by-step explanation:
This is what logarithms are for (among other things).
log(1.1^x) = log(100)
x·log(1.1) = 2
x = 2/log(1.1) ≈ 48.3177
Grandpa and Grandma are treating their family to the movies. Matinee tickets cost $4 per child and $4 per adult. Evening tickets cost $6 per child and $8 per adult. They plan on spending no more than $80 on the matinee tickets and no more than $100 on the evening tickets.
Complete Question
Grandpa and Grandma are treating their family to the movies. Matinee tickets cost $4 per child and $4 per adult. Evening tickets cost $6 per child and $8 per adult. They plan on spending no more than $80 on the matinee tickets and no more than $100 on the evening tickets. Could they take 9 children and 4 adults to both shows? Show your work. A yes or no answer is not sufficient for credit.
Answer:
Yes it is possible to take the 9 children and 4 adults to both shows
Step-by-step explanation:
From the question we are told that
The cost of the Matinee tickets for a child is z = $4
The cost of the Matinee tickets for an adult is a = $ 4
The cost of the Evening tickets for a child is k = $6
The cost of the Evening tickets for an adult is b = $8
The maximum amount to be spent on Matinee tickets is m = $80
The maximum amount to be spent on Evening tickets is e = $100
The number of child to be taken to the movies is n = 9
The number of adults to be taken to the movies is j = 4
Now the total amount of money that would be spent on Matinee tickets is mathematically evaluated as
[tex]t = 4 n + 4 j[/tex]
substituting values
[tex]t = 4 * 9 + 4* 4[/tex]
[tex]t = 52[/tex]
Now the total amount of money that would be spent on Evening ticket is mathematically evaluated as
[tex]T = 6n + 8j[/tex]
substituting values
[tex]T = 6(9) + 8(4)[/tex]
[tex]T = 86[/tex]
This implies that it is possible to take 9 children and 4 adults to both shows
given that
[tex]t \le m[/tex]
i.e $56 [tex]\le[/tex]$ 80
and
[tex]T \le e[/tex]
i.e $ 86 [tex]\le[/tex] $ 100
In 1998, as an advertising campaign, the Nabisco Company announced a "1000 Chips Challenge," claiming that every 18-ounce bag of their Chips Ahoy cookies contained at least 1000 chocolate chips. Dedicated statistics students at the Air Force Academy (no kidding) purchased some randomly selected bags of cookies and counted the chocolate chips. Some of their data are given below. 1219 1214 1087 1200 1419 1121 1325 1345 1244 1258 1356 1132 1191 1270 1295 1135 Find a 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies.
Answer:
A 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].
Step-by-step explanation:
We are given that statistics students at the Air Force Academy (no kidding) purchased some randomly selected bags of cookies and counted the chocolate chips.
Some of their data are given below; 1219, 1214, 1087, 1200, 1419, 1121, 1325, 1345, 1244, 1258, 1356, 1132, 1191, 1270, 1295, 1135.
Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean number of chocolate chips = [tex]\frac{\sum X}{n}[/tex] = 1238.2
s = sample standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }[/tex] = 94.3
n = sample of car drivers = 16
[tex]\mu[/tex] = population mean number of chips in a bag
Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.
So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.131 < [tex]t_1_5[/tex] < 2.131) = 0.95 {As the critical value of t at 15 degrees of
freedom are -2.131 & 2.131 with P = 2.5%}
P(-2.131 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.131) = 0.95
P( [tex]-2.131 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.131 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-2.131 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.131 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.131 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.131 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]1238.2-2.131 \times {\frac{94.3}{\sqrt{16} } }[/tex] , [tex]1238.2+2.131 \times {\frac{94.3}{\sqrt{16} } }[/tex] ]
= [1187.96, 1288.44]
Therefore, a 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].
Find the area of the yellow region.
Round to the nearest tenth.
15 cm
15 cm
Area = [ ? ] cm2
Answer:
48.3 cm²
Step-by-step explanation:
Let A be the area of the yellow region
A= the area of the square - the area of the quarter square
A= 15²-(15²*π)/4= 48.28≈ 48.3 cm²
Solve this correctly for brainliest !!!!!! 3(7) + 2 • |7 - 8| - 12
Answer:
3(7) + 2* |7 - 8| - 12 = 11
Step-by-step explanation:
3(7) + 2* |7 - 8| - 12
21 + 2* |-1| - 12
21 + 2* 1 - 12
21 + 2 - 12
23 - 12 = 11
Hope this helps! :)
Help please! Simplify 7/ √x
Answer:
[tex]\frac{7\sqrt{x} }{x}[/tex]
Step-by-step explanation:
To simplify 7/√x, we need to rationalize:
[tex]\frac{7}{\sqrt{x} } (\frac{\sqrt{x} }{\sqrt{x} } )[/tex]
When we multiply the 2, we should get our answer:
[tex]\frac{7\sqrt{x} }{x}[/tex]
Answer:
[tex]\frac{7\sqrt{x} }{x}[/tex]
Step-by-step explanation:
[tex]\frac{7}{\sqrt{x} } \\\\\frac{7}{\sqrt{x} } * \frac{\sqrt{x} }{\sqrt{x} } \\\\\frac{7\sqrt{x} }{\sqrt{x\sqrt{x} } } \\[/tex]
[tex]\frac{7\sqrt{x} }{x}[/tex]
Hope this helps! :)
Lisa drew three circles to form a figure. The areas of the circles were in the
ratio 1:4:16. She then shaded some parts of the figure as shown.
What fraction of the figure was shaded?
Answer:
Fraction of the figure shaded = [tex]\frac{13}{16}[/tex]
Step-by-step explanation:
Ratio of the areas of the given circles are 1 : 4 : 16
Then the radii of the circles will be in the ratio = [tex]\sqrt{1}:\sqrt{4}:\sqrt{16}[/tex]
= 1 : 2 : 4
If the radius of the smallest circle = x units
Then the radius of the middle circle = 2x units
and the radius of the largest circle = 4x units
Area of the smallest circle = πx²
Area of the middle circle = π(2x)² = 4πx²
Area of the largest circle = π(4x)²= 16πx²
Area of the region which is not shaded in the middle circle = πx²(4 - 1)
= 3πx²
Therefore, area of the shaded region = Area of the largest circle - Area of the region which is not shaded
= 16πx² - 3πx²
= 13πx²
Fraction of the figure which is not shaded = [tex]\frac{\text{Area of the shaded region}}{\text{Area of the largest circle}}[/tex]
= [tex]\frac{13\pi x^{2} }{16\pi x^{2} }[/tex]
= [tex]\frac{13}{16}[/tex]
The table shows three unique functions. (TABLE IN PIC) Which statements comparing the functions are true? Select three options. Only f(x) and h(x) have y-intercepts. Only f(x) and h(x) have x-intercepts. The minimum of h(x) is less than the other minimums. The range of h(x) has more values than the other ranges. The maximum of g(x) is greater than the other maximums.
Answer:
(A)Only f(x) and h(x) have y-intercepts.
(C)The minimum of h(x) is less than the other minimums.
(E)The maximum of g(x) is greater than the other maximums.
Step-by-step explanation:
From the table
f(0)=0 and h(0)=0, therefore, Only f(x) and h(x) have y-intercepts. (Option A)
Minimum of f(x)=-14Minimum of g(x)=1/49Minimum of h(x)=-28Therefore, the minimum of h(x) is less than the other minimums. (Option C).
Maximum of f(x)=14
Maximum of g(x)=49
Maximum of h(x)=0
Therefore, the maximum of g(x) is greater than the other maximums. (Option E)
Answer: It's B,C, and E
Step-by-step explanation:
What is the value of X in equation? 1/3 X - 2/3 = - 18
Answer:
x=-52
Step-by-step explanation:
1/3x=-17 1/3
x=-52
A very large batch of components has arrived at a distributor. The batch can be characterized as acceptable only if the proportion of defective components is at most .10. The distributor decides to randomly select 10 components and to accept the batch only if the number of defective components in the sample is at most 2. Let X denote the number of defective components in the sample. What is the distribution of X? Justify your answer.
Required:
What is the probability that the batch will be accepted when the actual proportion of defectives (p) is:_______
a, 0.01
b. 0.05
c. 0.10
d. 0.20
e. 0.25
Answer:
c. 0.10
Step-by-step explanation:
Hello!
To accept a batch of components, the proportion of defective components is at most 0.10.
X: Number of defective components in a sample of 10.
This variable has a binomial distribution with parameters n=10 and p= 0.10 (for this binomial experiment, the "success" is finding a defective component)
The distributor will accept the batch if at most two components are defective, symbolically:
P(X≤2)
Using the tables for the binomial distribution you can find the accumulated probability for a sample of n=10 with probability of success of p= 0.10 and number of successes x= 2
P(X≤2)= 0.9298
I hope this helps!
factorise 12x² + x - 20
━━━━━━━☆☆━━━━━━━
▹ Answer
(3x + 4) * (4x - 5)
▹ Step-by-Step Explanation
12x² + x - 20
Rewrite
12x² + 16x - 15x - 20
Factor out
4x(3x + 4) - 15x - 20
4x(3x + 4) - 5(3x + 4)
Factor
(3x + 4) * (4x - 5)
Hope this helps!
- CloutAnswers ❁
Brainliest is greatly appreciated!
━━━━━━━☆☆━━━━━━━
2) Find the diameter.
4) If the diameter is equal to 3 inches ,d=
Answer:
d = 3 in
Step-by-step explanation:
Since we are trying to find the diameter, and the diameter is given to us as 3 in, our diameter is 3 in.
is this right one more i think lol
Answer:
Yup P is the right one having 62.26%
Step-by-step explanation:
Answer:
Yes
Step-by-step explanation:
We can set it up as
P = 33/53
Q = 20/48
R = 54/90
S = 44/83
This is because we are calculating the percent of yellow birds in the total amt. of birds in a specified park.
Now we calculate =>
P = 33/53 = around 0.62
Q = 20/48 = around 0.416
R = 54/90 = 0.6
S = 44/83 = around 0.53
We find that Park P has the greatest percentage and -->
Thus, Park P is our answer and yes, you are correct.
what is u over 4-4= -20
u/4 - 4 = -20
Add 4 to both sides:
u/4 = -16
Multiply both sides by 4:
u = -64
Answer:
u=-64
Step-by-step explanation:
u/4 -4 = -20
First add 4 to both sides.
u/4=-16
Now multiply both sides by 4
u=-64
evaluate 25.1 * 2.51 in two decimal places
Answer:
63.00
Step-by-step explanation:
25.1 × 2.51
Multiply.
= 63.001
Round to two decimal places.
63.00
Answer:
63.00
Step-by-step explanation:
when u multiply 25.1 by 25.1 you get 630.01. Then u have to move the decimal over to the left once and then u get 63.00
Write a pair of integers whose sum is- -8
Answer:
-3+(-5)
Checking our answer:
Adding this does indeed give -8
Mario and tabitha are calculating the probability of getting a 4 and a 2 if they roll a die twice. Who is correct?
Answer:
[tex]\frac{2}{12}[/tex] simplified to [tex]\frac{1}{6}[/tex]
Step-by-step explanation:
4 = [tex]\frac{1}{12}[/tex]
2 = [tex]\frac{1}{12}[/tex]
[tex]\frac{1}{12}[/tex] + [tex]\frac{1}{12}[/tex] = [tex]\frac{2}{12}[/tex] ÷ 2 = [tex]\frac{1}{6}[/tex]
Using the following conversions between the metric and U.S. systems, convert the measurement. Round your answer to 6 decimal places as needed
1 meter ≈ 3.28 feet
1 Liter ≈ 0.26 gallons
1 kilogram ≈ 2.20 pounds
33.777 yd ≈ __________ km
Answer:
33.777 yd = 0.030886 km
Step-by-step explanation:
==>Given:
33.777 yd
==>Required:
Convert 33.777 yd to km to 6 decimal places, using the metric and U.S systems.
==>Solution:
To convert, note that 1 km = 1093.6133 yd.
Thus,
1 km = 1093.6133 yd
x km = 33.777 yd
Cross multiply
1 × 33.777 = 1093.6133 × x
33.777 = 1093.6133x
Divide both sides by 1093.6133, to solve for x
33.777/1093.6133 = x
0.03088569 = x
x ≈ 0.030886 (to 6 decimal places)
Therefore, 33.777 yd = 0.030886 km
Factor completely 5x(x + 3) + 6(x + 3). (1 point)
Answer:
The answer is ( 5x + 6 ) ( x + 3 )Step-by-step explanation:
5x(x + 3) + 6(x + 3)
The final answer is
( 5x + 6 ) ( x + 3 )
Hope this helps you
Which of the following functions is graphed below?
Answer:
C
Step-by-step explanation:
C is the solution
Answer:
Option C
Step-by-step explanation:
The graph is a horizontal translation 4 units left and a vertical translation 2 units down ⇒ y= |x+4|-2
find the value of x that makes abcd a parallelogram
The 4 angles need to add to 360.
2 of them are 70
The other two need to equal 360-140 = 220
They are both the same so one angle needs to equal 220/2 = 110
Now find x:
X + 60 = 110
Subtract 60 from both sides:
X = 50. The answer is D
A square has a perimeter of 12x+52 units. Which expression represents the side leagth of the square in units
Answer:
12x/2 or 52/2
Step-by-step explanation:
Ok, perimeter is length+length+width+width. 12x/2 and 52/2 could are probably the answers.
An article gave the accompanying data on ultimate load (kN) for two different types of beams. Assuming the underlying distributions are Normal, calculate and interpret a 99% Cl for the difference between the true average load for the fiberglass beams and that for the carbon beams.
Type Sample size Sample Mean Sample SD
Fiberglass grid 26 33.4 2.2
Commercial carbon 26 42.8 4.3
grid
1. Calculate and interpret a 99% Cl for true average stance duration among elderly individuals.
2. Carry out a test of hypotheses at significance level 0.05 to decide whether true average stance duration is larger among elderly individuals than younger individuals.
Answer:
The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is (-11.937, -6.863).
Step-by-step explanation:
We have to calculate a 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams.
The sample 1 (Fiberglass), of size n1=26 has a mean of 33.4 and a standard deviation of 2.2.
The sample 2 (Carbon), of size n2=26 has a mean of 42.8 and a standard deviation of 4.3.
The difference between sample means is Md=-9.4.
[tex]M_d=M_1-M_2=33.4-42.8=-9.4[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{2.2^2}{26}+\dfrac{4.3^2}{26}}\\\\\\s_{M_d}=\sqrt{0.186+0.711}=\sqrt{0.897}=0.9473[/tex]
The critical t-value for a 99% confidence interval is t=2.678.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_{M_d}=2.678 \cdot 0.9473=2.537[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M_d-t \cdot s_{M_d} = -9.4-2.537=-11.937\\\\UL=M_d+t \cdot s_{M_d} = -9.4+2.537=-6.863[/tex]
The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is (-11.937, -6.863).
In this way, we can calculate the individual duration of each one and the duration time, knowing that the sample means:
The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is -11.937 and -6.863.
We have to calculate a 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams. The sample 1 (Fiberglass), of size n1=26 has a mean of 33.4 and a standard deviation of 2.2. The sample 2 (Carbon), of size n2=26 has a mean of 42.8 and a standard deviation of 4.3. The difference between sample means is Md=-9.4.
[tex]Sm_d= \sqrt{\frac{\sigma^2_1}{n_1} +\frac{\sigma^2_2}{n_2}} = \sqrt{(0.186)+(0.711) }= 0.9473[/tex]
The critical t-value for a 99% confidednce interval is t=2.678. The margin of error (MOE) can be calculated as:
[tex]MOE=t*8M_d = (2.678)(0.9473)= 2.537[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL= M_d-t*SM_d = -9.4-2.537= -11.937\\UL= M_d+t*SM_d= -9.4+2.537= -6.863[/tex]
The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is (-11.937, -6.863).
See more about statistics at brainly.com/question/2289255
0.3y+ z y 0, point, 3, y, plus, start fraction, y, divided by, z, end fraction when y=10y=10y, equals, 10 and z=5z=5z, equals, 5.
Answer:
5
Step-by-step explanation:
Substitute the given values and do the arithmetic.
[tex]0.3y+\dfrac{y}{z}=0.3\cdot 10+\dfrac{10}{5}=3+2=\boxed{5}[/tex]
What is the justification for step 2 in the solution process?
Answer:
Answer C
Step-by-step explanation:
You are balancing this equation out by subtracting 7x from both sides. This means you are using the subtraction property of equality.
what is 3 + 3 × 3 + 3 =
Answer:
15
Step-by-step explanation:
PEMDAS
3x3 = 9
3+3 = 6
9+6 = 15
By the BODMAS rule we get, 3 + 3 × 3 + 3 = 15
The acronym BODMAS rule is used to keep track of the right sequence of operations to do when solving mathematical issues. Brackets (B), order of powers or roots (O), division (D), multiplication (M), addition (A), and subtraction (S) are all represented by this acronym (S).
3 + 3 × 3 + 3 =
3 × 3 = 9
3 + 9 + 3 = 15.
Therefore, the correct answer is 15.
Learn more about BODMAS rule here:
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Which number is greatest? 6.23 times 10 Superscript 12 6.23 times 10 Superscript 8 6.23 times 10 Superscript negative 6 6.23 times 10 Superscript 3
The greatest number is 6.23 times 10 superscript 12.
How does scientific notations work?The number is written in the form [tex]a \times 10^b[/tex] where we have [tex]1 \leq a < 10[/tex]
The number b shows the order, which is the most important figure for which scientific notation is used. It tells us how much order large or small a value is in powers of 10. We can for a time, ignore the value of 'a' for two comparable quantities and only compare their orders(this type of comparison is useful when difference is too big, like size of human to size of a star etc sort of comparisons).
We are given that the number so;
A.6.23 x 10^12 is equivalent to rolling the decimal 12 times to the right.
B.6.23 x 10^8 is equivalent to rolling the decimal 8 times to the right.
C.6.23 x 10^-6 is equivalent to rolling the decimal 6 times to the left.
D.6.23 x 10^3 is equivalent to rolling the decimal 3 times to the right.
This shows the 10 has been multiplied by itself thrice.
Learn more about scientific notation here:
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Dairy cows at large commercial farms often receive injections of bST (Bovine Somatotropin), a hormone used to spur milk production. Bauman et al. (Journal of Dairy Science, 1989) reported that 12 cows given bST produced an average of 28.0 kg/d of milk. Assume that the standart deviation of milk production is 2.25 kg/d.
Requried:
a. Find a 99% confidence interval for the true mean milk production.
b. If the farms want the confidence interval to be no wider than ± 1.25 kg/d, what level of confidence would they need to use?
Answer:
a) 26.33 kg/d and 29.67 kg/d
b) 94.5%
Step-by-step explanation:
a. Find a 99% confidence interval for the true mean milk production.
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 2.575*\frac{2.25}{\sqrt{12}} = 1.67[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 28 - 1.67 = 26.33 kg/d
The upper end of the interval is the sample mean added to M. So it is 28 + 1.67 = 29.67 kg/d
The 99% confidence interval for the true mean milk production is between 26.33 kg/d and 29.67 kg/d
b. If the farms want the confidence interval to be no wider than ± 1.25 kg/d, what level of confidence would they need to use?
We need to find z initially, when M = 1.25.
[tex]M = z*\frac{2.25}{\sqrt{12}} = 1.67[/tex]
[tex]1.25 = z*\frac{2.25}{\sqrt{12}} = 1.67[/tex]
[tex]2.25z = 1.25\sqrt{12}[/tex]
[tex]z = \frac{1.25\sqrt{12}}{2.25}[/tex]
[tex]z = 1.92[/tex]
When [tex]z = 1.92[/tex], it has a pvalue of 0.9725.
1 - 2*(1 - 0.9725) = 0.945
So we should use a confidence level of 94.5%.
Dr. Pagels is a mammalogist who studies meadow and common voles. He frequently traps the moles and has noticed what appears to be a preference for a peanut butter-oatmeal mixture by the meadow voles vs apple slices are usually used in traps, where the common voles seem to prefer the apple slices. So he conducted a study where he used a peanut butter-oatmeal mixture in half the traps and the normal apple slices in his remaining traps to see if there was a food preference between the two different voles.
Indicate which of the following is the null hypothesis, and which is the alternate hypothesis.
There food preferences among vole species are independent of one another. _____
There is a relationship between voles and food preference. ______
To test for independence, we need to calculate the Chi-square statistic.
These are the data that Dr. Pagels collected:
meadow voles common voles
apple slices 26 32
peanut butter-oatmeal 35 25
When transferring your answers, make sure you carry them out to AT LEAST SIX SIGNIFICANT FIGURES unless otherwise stated.
_____= expected meadow vole/apple slices
_____= expected common vole/apple slices
_____= expected meadow vole/peanut butter-oatmeal
_____= expected common vole/peanut butter-oatmeal
_____= chi-square value
_____= degrees of freedom (whole number only)
_____= using Statistical Table A (pg 704 of your textbook), what is the chi-square critical value with significance level of alpha=0.05?
_____= will you reject or fail to reject the null hypothesis? (answer either reject or fail to reject)
Answer:
Null hypothesis = H₀ = There food preferences among vole species are independent of one another.
Alternate hypothesis = H₁ = There is a relationship between voles and food preference.
Expected meadow vole/apple slices = 29.983051
Expected common vole/apple slices = 28.016949
Expected meadow vole/peanut butter-oatmeal = 31.016949
Expected common vole/peanut butter-oatmeal = 28.983051
Chi-square value = χ² = 2.154239
Degree of freedom = 1
Critical value = 3.841
χ² < Critical value
We failed to reject H₀
We do not have significant evidence at the given significance level to show that there is a relationship between voles and food preference.
Step-by-step explanation:
He frequently traps the moles and has noticed what appears to be a preference for a peanut butter-oatmeal mixture by the meadow voles vs apple slices are usually used in traps, where the common voles seem to prefer the apple slices.
So he conducted a study where he used a peanut butter-oatmeal mixture in half the traps and the normal apple slices in his remaining traps to see if there was a food preference between the two different voles.
Null hypothesis = H₀ = There food preferences among vole species are independent of one another.
Alternate hypothesis = H₁ = There is a relationship between voles and food preference.
Data collected by Dr. Pagels:
meadow voles common voles Row Total
apple slices 26 32 58
peanut butter-oatmeal 35 25 60
Column Total 61 57 118
Where 118 is the grand total.
The expected number is given by
Expected = (row total)×(column total)/grand total
Expected meadow vole/apple slices = 58×61/118
Expected meadow vole/apple slices = 29.983051
Expected common vole/apple slices = 58×57/118
Expected common vole/apple slices = 28.016949
Expected meadow vole/peanut butter-oatmeal = 60×61/118
Expected meadow vole/peanut butter-oatmeal = 31.016949
Expected common vole/peanut butter-oatmeal = 60×57/118
Expected common vole/peanut butter-oatmeal = 28.983051
The chi-square statistic value is given by
χ² = Σ(Observed - Expected)²/Expected
χ² = (26 - 29.983051)²/29.983051 + (32 - 28.016949)²/28.016949 + (35 - 31.016949)²/31.016949 + (25 - 28.983051)²/28.983051
χ² = 2.154239
The degrees of freedom is given by
DoF = (row - 1)×(col - 1)
For the given case, we have 2 rows and 2 columns
DoF = (2 - 1)×(2 - 1)
DoF = 1
The given level of significance = 0.05
The critical value from the chi-square table at α = 0.05 and DoF = 1 is found to be
Critical value = 3.841
Conclusion:
Reject H₀ If χ² > Critical value
We reject the Null hypothesis If the calculated chi-square value is more than the critical value.
For the given case,
χ² < Critical value
We failed to reject H₀
We do not have significant evidence at the given significance level to show that there is a relationship between voles and food preference.