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2 years ago

No chi-square The principal in Exercise 7 also

Mathematics

1 answer:

larisa [96]2 years ago

8 0

The random sample of the students is an illustration of sampling

The chi-square test for goodness of fit is inappropriate because the variable under study is not categorical.

<h3>How to determine the reason chi square is not appropriate?</h3>

The dataset is given as:

Monday 34

Tuesday 29

Wednesday 32

Thursday 28

Friday 19

The variable of the above dataset is a not a categorical dataset.

One of the conditions of the chi-square test for goodness of fit test is that the variable under study must be categorical.

Hence, the chi-square test for goodness of fit is inappropriate because the variable under study is not categorical.

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Evaluate f(4) f(x) = 3x + 4

RSB [31]

The answer is definitely 7xf

7 0

3 years ago

Amy hikes down a slope to a lake that is 10.2 meters below the trail. Then Amy jumps into the lake, and swims 1.5 meters down. S

Grace [21]

she went down the slope 10.2 meters, -10.2

then she jumped down to the lake 1.5 meters, -1.5

-10.2 - 1.5 = -11.7 meters.

so she pretty much went 11.7 meters down from her original location.

7 0

3 years ago

Read 2 more answers

Y = 6x + 2 y = -6x + 2 These lines are?

77julia77 [94]

Answer:parallel

Step-by-step explanation:

they will be parallel because they are equal

5 0

3 years ago

Advertisements for the Sylph Physical Fitness Center claim that completion of their course will result in a loss of weight (meas

MrMuchimi

Answer:

90% confidence interval for the mean weight loss for the population represented by this sample assuming that the differences are coming from a normally distributed population is [1.989 ,13.5105]

Step-by-step explanation:

The data given is

Mean Std Dev

Before 177.25 29.325

After 169.50 22.431

Difference 7.75 8.598

Hence d`= 7.75 and sd= 8.598

The 90% confidence interval for the difference in means for the paired observation is given by

d` ± t∝/2(n-1) *sd/√n

Here t∝/2(n-1)=1.895 where n-1= 8-1= 7 d.f

and ∝/2= 0.1/2=0.05

Putting the values

d` ± t∝/2(n-1) *sd/√n

7.75 ±1.895 * 8.598 /√8

7.75 ± 5.7605

1.989 ,13.5105

90% confidence interval for the mean weight loss for the population represented by this sample assuming that the differences are coming from a normally distributed population is [1.989 ,13.5105]

3 0

3 years ago

An animal shelter houses dogs, cats, and rabbits. There are 126 animals at the shelter. Of the animals, 1/3 are cats. Three four

navik [9.2K]

Answer:

21 rabbits

Step-by-step explanation:

⇒ Given:

There are 126 animals at the shelter.

Of the animals, 1/3 are cats.

Three fourths of the remaining animals are dogs.

To Find:

How many of the animals are rabbits? Show your work.

Solve:

Since 1/3 are cats and the total is 126.

Then, 1/3 of 126 is 42. {1/3 × 126 = 42}

Therefore, there are 42 cats.

3/4 of the remaining animals, which is 84 are dogs.

Which 3/4 of 84 is 63. { 3/4 × 84 = 63 }

So there are 63 dogs.

Now adding dogs and cats together:

63 + 42 = 105 animals.

126 {Total} - 105{cat and cats} = 21 {rabbits}

Hence, there are 21 rabbits.

Kavinsky

7 0

2 years ago