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3 years ago

The z value for a 97. 8% confidence interval estimation is.

Mathematics

1 answer:

VARVARA [1.3K]2 years ago

4 0

Using the z-distribution, it is found that the z-value is of z = -2.29.

<h3>How to find the critical value of Z?</h3>

For a confidence level of $\alpha$ , it is Z with a p-value of $\frac{1 + \alpha}{2}$ .

We have a 97.8% confidence level, hence $\alpha = 0.978$ , z is the value of Z that has a p-value of $\frac{1+0.978}{2} = 0.989$ , so the critical value is z = -2.29.

More can be learned about the z-distribution at brainly.com/question/25890103

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astra-53 [7]

Answer:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference of birthdates distributed throughout the year

H1: There is a difference between birthdates distributed throughout the year

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total}{4}$

And replacing we got:

$E_{1} =\frac{67+56+30+37}{4}=47.5$

And now we can calculate the statistic:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

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4 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

pristine printing will print business cards for $0.10 each plus a setup charge of $15 and printing place offers business cards f

Luden [163]

Let's assume

total number of cards =x

Pristine printing business cards:

for $0.10 each plus a setup charge of $15

so, we get equation for cost as

$C=15+0.10x$

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for $0.15 each with a setup charge of $10

so, we get equation for cost as

$C=10+0.15x$

now, we are given

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4 years ago

X^2 +105x-1050=1550 please help

bazaltina [42]

-x^2+105x-1050=1550
We move all terms to the left:
-x^2+105x-1050-(1550)=0
We add all the numbers together, and all the variables
-1x^2+105x-2600=0
a = -1; b = 105; c = -2600;
Δ = b2-4ac
Δ = 1052-4·(-1)·(-2600)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
x1=−b−Δ√2ax2=−b+Δ√2a
Δ−−√=625−−−√=25 x1=−b−Δ√2a=−(105)−252∗−1=−130−2=+65 x2=−b+Δ√2a=−(105)+252∗−1=−80−2=+40

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4 years ago

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Answer:

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Step-by-step explanation:

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