Write in standard form. (Use Desmos)* 9.76x10 with a 5 on top of the ten

A zoologist is studying four very closely related feline species. She wishes to compare their gestation periods. An observationa

diamong [38]

Answer:

$p_v= 0.01$

Since the significance level is 0.05 we see that $pv$ so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different, NOT to conclude that the all the means are different.

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

Solution to the problem

The hypothesis for this case are:

Null hypothesis: $\mu_{A}=\mu_{B}=\mu_{C}= \mu_D$

Alternative hypothesis: Not all the means are equal $\mu_{i}\neq \mu_{j}, i,j=A,B,C,D$

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have $p=4$ groups and on each group from $j=1,\dots,p$ we have $n_j$ individuals on each group we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$

And we have this property

$SST=SS_{between}+SS_{within}$

And in order to test this hypothesis we need to ue an F statistic and for this case the p value calculated is

$p_v= 0.01$

Since the significance level is 0.05 we see that $pv$ so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different NOT to conclude that the all the means are different.

8 0

3 years ago

A warehouse has 784 barrels of a chemical. You require 98 barrels for one run of your process. How many barrels will be left aft

sergejj [24]

686 barrels left. If I'm not mistaken this should just be simple subtraction

4 0

3 years ago

11. Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x) 5 5 0 x , 0

NISA [10]

Question not properly presented

Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x)

0 ------ x<0

x²/25 ---- 0 ≤ x ≤ 5

1 ----- 5 ≤ x

Use the cdf to obtain the following.

(a) Calculate P(X ≤ 4).

(b) Calculate P(3.5 ≤ X ≤ 4).

(c) Calculate P(X > 4.5)

(d) What is the median checkout duration, μ?

e. Obtain the density function f (x).

f. Calculate E(X).

Answer:

a. P(X ≤ 4) = 16/25

b. P(3.5 ≤ X ≤ 4) = 3.75/25

c. P(4.5 ≤ X ≤ 5) = 4.75/25

d. μ = 3.5

e. f(x) = 2x/25 for 0≤x≤2/5

f. E(x) = 16/9375

Step-by-step explanation:

a. Calculate P(X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(X ≤ 4) = F(x) {0,4}

P(X ≤ 4) = x²/25 {0,4}

P(X ≤ 4) = 4²/25

P(X ≤ 4) = 16/25

b. Calculate P(3.5 ≤ X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(3.5 ≤ X ≤ 4) = F(x) {3.5,4}

P(3.5 ≤ X ≤ 4) = x²/25 {3.5,4}

P(3.5 ≤ X ≤ 4) = 4²/25 - 3.5²/25

P(3.5 ≤ X ≤ 4) = 16/25 - 12.25/25

P(3.5 ≤ X ≤ 4) = 3.75/25

(c) Calculate P(X > 4.5).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(4.5 ≤ X ≤ 5) = F(x) {4.5,5}

P(4.5 ≤ X ≤ 5) = x²/25 {4.5,5}

P(4.5 ≤ X ≤ 5)) = 5²/25 - 4.5²/25

P(4.5 ≤ X ≤ 5) = 25/25 - 20.25/25

P(4.5 ≤ X ≤ 5) = 4.75/25

(d) What is the median checkout duration, μ?

Median is calculated as follows;

∫f(x) dx {-∝,μ} = ½

This implies

F(x) {-∝,μ} = ½

where F(x) = x²/25 for 0 ≤ x ≤ 5

F(x) {-∝,μ} = ½ becomes

x²/25 {0,μ} = ½

μ² = ½ * 25

μ² = 12.5

μ = √12.5

μ = 3.5

e. Calculating density function f (x).

If F(x) = ∫f(x) dx

Then f(x) = d/dx (F(x))

where F(x) = x²/25 for 0 ≤ x ≤ 5

f(x) = d/dx(x²/25)

f(x) = 2x/25

When

F(x) = 0, f(x) = 2(0)/25 = 0

When

F(x) = 5, f(x) = 2(5)/25 = 2/5

f(x) = 2x/25 for 0≤x≤2/5

f. Calculating E(X).

E(x) = ∫xf(x) dx, 0,2/5

E(x) = ∫x * 2x/25 dx, 0,2/5

E(x) = 2∫x ²/25 dx, 0,2/5

E(x) = 2x³/75 , 0,2/5

E(x) = 2(2/5)³/75

E(x) = 16/9375

4 0

3 years ago

20 points free<br> How to make a cow bark

yKpoI14uk [10]

Answer:

RUFF RUFF RUFF RUFF RUFF

6 0

3 years ago

Janelle has a 12 ounce can of soft drink. She drink 3/5 of the soft drink. How many ounces of soft drink are left in tbe can?

ivann1987 [24]

Answer: 4.8 ounces

Step-by-step explanation:

In total there are 12 ounces. If she drinks 3/5 of 12 you multiply 3/5 by 12.

3/5 * 12 would be equal to 3/5 * 12/1 which is 36/5. You have found how much she has drank. 36/5 is also equal to 7.2 ounces. Subtract 7.2 form 12 and you get 4.8 ounces.

Hope this helped!

7 0

3 years ago

Write in standard form. (Use Desmos)* 9.76x10 with a 5 on top of the ten​

Write in standard form. (Use Desmos)* 9.76x10 with a 5 on top of the ten