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damaskus [11]
3 years ago
13

Suppose a researcher is testing the hypothesis Upper H 0​: pequals0.4 versus Upper H 1​: pless than0.4 and she finds the​ P-valu

e to be 0.33. Explain what this means. Would she reject the null​ hypothesis? Why? Choose the correct explanation below. A. If the​ P-value for a particular test statistic is 0.33​, she expects results no more extreme than the test statistic in about 33 of 100 samples if the null hypothesis is true. B. If the​ P-value for a particular test statistic is 0.33​, she expects results no more extreme than the test statistic in exactly 33 of 100 samples if the null hypothesis is true. C. If the​ P-value for a particular test statistic is 0.33​, she expects results at least as extreme as the test statistic in about 33 of 100 samples if the null hypothesis is true. D. If the​ P-value for a particular test statistic is 0.33​, she expects results at least as extreme as the test statistic in exactly 33 of 100 samples if the null hypothesis is true. Choose the correct conclusion below. A. Since this event is​ unusual, she will reject the null hypothesis. B. Since this event is not​ unusual, she will reject the null hypothesis. C. Since this event is​ unusual, she will not reject the null hypothesis. D. Since this event is not​ unusual, she will not reject the null hypothesis.
Mathematics
1 answer:
andre [41]3 years ago
3 0

Answer:

D. If the​ P-value for a particular test statistic is 0.33​, she expects results at least as extreme as the test statistic in exactly 33 of 100 samples if the null hypothesis is true.

D. Since this event is not​ unusual, she will not reject the null hypothesis.

Step-by-step explanation:

Hello!

You have the following hypothesis:

H₀: ρ = 0.4

H₁: ρ < 0.4

Calculated p-value: 0.33

Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

In this case, you have a 33% chance of getting a value as extreme as the statistic value if the null hypothesis is true. In other words, you would expect results as extreme as the calculated statistic in 33 about 100 samples if the null hypothesis is true.

You didn't exactly specify a level of significance for the test, so, I'll use the most common one to make a decision: α: 0.05

Remember:

If p-value ≤ α, then you reject the null hypothesis.

If p-value > α, then you do not reject the null hypothesis.

Since 0.33 > 0.05 then I'll support the null hypothesis.

I hope it helps!

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Define z_alpha to be a z-score with an area of alpha to the right. For Example: z_0.10 means P(Z &gt; z_0.10) = 0.10. We would a
Reptile [31]

Answer:

a) P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

"=NORM.INV(0.025,0,1)"

And for this case the two values are :z_{crit}= \pm 1.96

b) P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates \alpha/2 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

"=NORM.INV(alpha/2,0,1)"

c) For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got z_{\alpha/2}=1.64

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Part a

P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

"=NORM.INV(0.025,0,1)"

And for this case the two values are :z_{crit}= \pm 1.96

Part b

P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates \alpha/2 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

"=NORM.INV(alpha/2,0,1)"

Part c

For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got z_{\alpha/2}=1.64

6 0
3 years ago
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