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HACTEHA [7]
3 years ago
6

In a​ one-tail hypothesis test where you reject Upper H0 only in the lower ​tail, it was found that the p-value is 0.0256 if ZST

AT = -1.95. What is your statistical decision if you test the null hypothesis at the 0.10 level of​ significance?
A) Reject the null hypothesis because the​ p-value is greater than the level of significance.B) Reject the null hypothesis because the​ p-value is less than the level of significanceC) Fail to reject the null hypothesis because the​ p-value is less than the level of significance.D) Fail to reject the null hypothesis because the​ p-value is greater than the level of significance
Mathematics
1 answer:
kow [346]3 years ago
7 0

Answer:

B) Reject the null hypothesis because the​ p-value is less than the level of significance

Step-by-step explanation:

Given that in a one-tail hypothesis test where you reject Upper H0 only in the lower ​tail, it was found that the p-value is 0.0256 if ZSTAT = -1.95.

So p value = 0.0256

But alpha here is 0.10

Since p <0.10

Whenever p is less than alpha, we reject null hypothesis

Hence correct answer is

B) Reject the null hypothesis because the​ p-value is less than the level of significance

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3 years ago
The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the f
Verizon [17]

Answer:

CI(95%): [205,5;400.8]dolars

Step-by-step explanation:

Hello!

So you need to construct a confidence interval for the average family dental expenses (μ) using the sample given in the problem. To estimate it you need to first choose a statistic. For this small sample, considering that the variable has a normal distribution (I made a quick Shapiro Wilks test, with p-value 0.3234, you can assume normality) the best statistic to use is the Student's t-test.

t= [x(bar)-μ]/S/√n ≈ t₍ₙ₋₁₎

The formula for the confidence interval to estimate the mean is

x(bar)±t_{n-1; 1-\alpha/2}* (S/√n)

<u>The critical value is from a t-distribution with 11 degrees of freedom </u>

±t_{n-1; 1-\alpha/2} = t_{11; 0.975} = 2.201

<em> >remember since it's two-tailed, to get the right critical value you have to divide α by 2. So in the text, you received a confidence level of 1-α=0.95 so α=0.05 then α/2=0.025 and 1-α/2=0.975</em>

To construct the interval, you need to first calculate the sample mean and the standard derivation.

<u>Sample</u>

115; 370; 250; 593; 540; 225; 117; 425; 318; 182; 275; 228

n= 12

∑xi = 3638 ∑xi² = 1362670

<u>Sample mean</u>

x(bar): (∑xi)/n = 3638/12 = 303.17 dolars

<u>Standard derivation</u>

S²= 1/n-1*[∑xi²- (∑xi)²/n] = 1/11 * [1362670-((3638)²/12)] = 23614.47 dolars²

S= 153.67 dolars

<u>Confidence interval (95%)</u>

303.17± 2.201* (153.67/√12)

[205,5;400.8]dolars

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What is the x-value at which the graphs of y = x2 – 2x + 1 and y = x2 + 2x – 7 inter-<br> sect?
Kitty [74]

Answer:

x = 2

Step-by-step explanation:

Both equations are equal to y, so they're also equal to each other.  We then set them equal to each other:

x^2 - 2x + 1  =  x^2 + 2x - 7

We now do algebra to isolate x.  Subtract 1 from both sides.

x^2 - 2x  =  x^2 + 2x - 8

Subtract 2x from both sides.

x^2 - 4x  =  x^2 - 8

Subtract x^2 from both sides.

-4x = -8

Divide both sides by -4.

x = 2

5 0
3 years ago
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