Learn Significance Test with Examples & Questions

What are the characteristics of a good hypothesis?

The numbers of significant figures in $9.1\times <!--\; \times \; -->{10}^{-<!--\; -\; -->31}kg$ are:
A)Two
B)Three
C)Ten
D)Thirty one

What is the five-step process for hypothesis testing?

How to calculate Type 1 error and Type 2 error probabilities?

What is the square root of 106?

Perform hypothesis test for population mean It is claimed that average rice production by Company2 is 209,500 cwt (Company2 = 209500.0). Test this claim using a hypothesis test at 1% level of significance.
My answer: Null = 209500.0 alternative = .01
I'm being told that these my answers are not correct. I have no idea why. Any help would be appreciated!

If you are given a population of students doing hypothesis tests for a certain condition at a certain significance level, is it possible to calculate:
(a) how many students will fail to reject the null hypothesis given that the null hypothesis is false
(b) how many students will reject the null hypothesis given that the null hypothesis is true.
I have tried searching online but so far all sites only show how to calculate the probability that at least one type I error will be made. Any assistance will be greatly appreciated.

Why is standard deviation calculated differently for finding Z scores and confidence intervals
Suppose that as a personnel director, you want to test the perception of fairness of two methods of performance evaluation. 63 of 78 employees rated Method 1 as fair. 49 of 82 rated Method 2 as fair. A 99% confidence interval for $p_1$−$p_2$ (where $p_1$ = 63/78 and $p_2$ = 49/82) is as follows:
$p_1-<!--\; -\; -->p_2\pm <!--\; \pm \; -->2.58\sqrt{\frac{p_1(1-<!--\; -\; -->p_1)}{n_1}+\frac{p_2(1-<!--\; -\; -->p_2)}{n_2}}$
$0.029⩽<!--\; ⩽\; -->p_1-<!--\; -\; -->p_2⩽<!--\; ⩽\; -->0.391$
At the 0.01 level of significance the Z score is
$Z=\sqrt{\frac{p(1-<!--\; -\; -->p)}{n_1}+\frac{p(1-<!--\; -\; -->p)}{n_2}}$
(where p = $(x_1+x_2)/(n_1+n_2)=0.70$ but sometimes a different formula $p=(p_1+p_2)/2$ is also used)
Z = 2.90
Both tests indicate that there is evidence of a difference.
But you could also find the Z score using the standard deviation formula in the first method to be 2.993. Why are the Z scores different? Where do the formulas for finding the standard deviation come from?

Suppose $X_n$ is an i.i.d. random sample from the $N(\mu ,{\sigma }^2)$ population, where μ is unknown but ${\sigma }^2$ is known. Consider a test statistic $T=\sqrt{n}{(}^{\prime }bar{X}_n-{\mu }_0)/\sigma$ at the significance level $\mathrm{\alpha <!--\; \alpha \; -->}$ for $H_0:\mu ={\mu }_{0}versus{H}_a:\mu \ne <!--\; \ne \; -->{\mu }_0$. Find Type I error of this test.
Confusion:
One of my friends gives this answer:
The hypothesis model will be
$H_o:\mathrm{\mu <!--\; \mu \; -->}={\mathrm{\mu <!--\; \mu \; -->}}_o$
$H_A:\mathrm{\mu <!--\; \mu \; -->}\ne <!--\; \ne \; -->{\mathrm{\mu <!--\; \mu \; -->}}_o$
Type I error $\mathrm{\alpha <!--\; \alpha \; -->}$ is the probablity of rejecting the null hypothesis $H_o$
Thus $p=\mathrm{\alpha <!--\; \alpha \; -->}$
I think it's not correct, because Type I error is the the probablity of rejecting the null hypothesis $H_o$ when $H_o$ is true , and α means the size of test is no larger than $\mathrm{\alpha <!--\; \alpha \; -->}$, so I think there are some differences between these two concepts.
So who is wrong? why? and what the correct answer should be?

You have a coin and p is the probability of heads.
You throw the coin until you obtain heads for the first time.
You want to test $H_0:p=1/2$ against $H_1:p<1/2$. You reject $H_0$ in favor of $H_1$ if $T\ge k$, where T is the number of the throw which yielded heads for the first time and k an integer.
Determine the smallest value of k corresponding to the level of significance α=0.01.
Answer k=8
I can't find the correct solution for this question, how it's possible to have $H_1:p<1/2$ and then they say that we reject $H_0$ in favor of $H_1$ if $T\ge k$, shouldn't they be in the same direction?
I have to compute $P(T>k|H_0)=P(T>k|p=1/2)$ right? How can I compute it?

A chi square test is conducted to check whether a person's ability in Mathematics has an impact on his/her interest in Statistic.
The test statistic is 13.277 under the tested null hypothesis. write a recommended null hypothesis and an alternative hypothesis. Briefly describe your conclusion on this test at the 0.01 significance level.

check the null hypothesis that the mean is extra than or identical to 50. Use a one tail and a importance level of .05.
A sample of 25 answers has an average=forty five and a variance=25.

Given there's a number generator which creates irregular numbers to 9. Given after 50 number eras as it were 5 of the created numbers are underneath 3 (i.e. 0,1 or 2) and the importance level is α = 1%. How can I test in the event that for this importance level the number generator is producing all numbers with equal probability? It is obvious to me that in the event that the number generator would be without a doubt producing the numbers with uniform likelihood, one would anticipate 15 created numbers between and 2 - but how do I test the nullhypothesis in this case accurately?

We have the Random Variable X, which is $\mathrm{\Gamma <!--\; \Gamma \; -->}(p,\mathrm{\lambda <!--\; \lambda \; -->})$ distributed with the density:
$$f_{p,\mathrm{\lambda <!--\; \lambda \; -->}}(x)=\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^p}{\mathrm{\Gamma <!--\; \Gamma \; -->}(p)}\cdot <!--\; \cdot \; -->x^{p-<!--\; -\; -->1}\cdot <!--\; \cdot \; -->e^{-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}x}$$
with p=10 and $H_0:\mathrm{\lambda <!--\; \lambda \; -->}=2$ or $H_1:\mathrm{\lambda <!--\; \lambda \; -->}=4$ and $\mathrm{\alpha <!--\; \alpha \; -->}=0.001$
I want to apply the Lemma of Neyman Pearson which states:
Be c>0 fixed and chosen in the way that $A(c)=\{x\in <!--\; \in \; -->B:\frac{f_0(x)}{f_1(x)}\ge <!--\; \ge \; -->c\}$ such that ${\mathbb{P}}_{H_0}(X\in <!--\; \in \; -->A(c))=\mathrm{\alpha <!--\; \alpha \; -->}$
Then the test with the region A(c) among all tests with significance level $\mathrm{\alpha <!--\; \alpha \; -->}$ is the most powerful.
I am now trying to calculate A(c), but got stuck. I have:
$${\int <!--\; \int \; -->}_{A(c)}{f}_0(x)dx={\int <!--\; \int \; -->}_{A(c)}\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^p}{\mathrm{\Gamma <!--\; \Gamma \; -->}(p)}\cdot <!--\; \cdot \; -->x^{p-<!--\; -\; -->1}\cdot <!--\; \cdot \; -->e^{-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}x}dx=\mathrm{\alpha <!--\; \alpha \; -->}.$$
But I don't know how to get A(c) from this integral...
$\frac{f_0(x)}{f_1(x)}=\frac{1}{1024}\cdot <!--\; \cdot \; -->e^{2x}$

Is P value Type I error in hypothesis testing?
I'm confused about the interpretation of P value in hypothesis testing. I know that we set significance level as 0.05 which is the threshold we set for this test so that it won't suffer from Type I error by 5%.
And we are comparing P to significance level, does it mean P is the probability of making type I error based on the sample?

Why "bother" with a null hypothesis at all?
whenever i am seeking to get into data (once more), i am constantly lost at speculation trying out.
My simple question is - why can we form a null hypothesis as a negation of what we need to show within the first region, and most effective then can we show or disprove the null speculation?
Why do we do it at all, in preference to just proving the authentic speculation?

A poll from a previous year showed that 10% of smartphone owners relied on their data plan as their primary form of internet access. Researchers were curious if that had changed, so they tested $H_0:p=10\mathrm{\%<!--\; \%\; -->}$ versus $H_a:p\ne <!--\; \ne \; -->10\mathrm{\%<!--\; \%\; -->}$ where p is the proportion of smartphone owners who rely on their data plan as their primary form of internet access. They surveyed a random sample of 500 smartphone owners and found that 13% of them relied on their data plan.
The test statistic for these results was $z\approx <!--\; \approx \; -->2.236$, and the corresponding P-value was approximately 0.025.
Assuming the conditions for inference were met, which of these is an appropriate conclusion?
a) At the $\mathrm{\alpha <!--\; \alpha \; -->}$=0.01 significance level, they should conclude that the proportion has changed from 10%.
b) At the $\mathrm{\alpha <!--\; \alpha \; -->}$=0.01 significance level, they should conclude that the proportion is still 10%.
c) At the $\mathrm{\alpha <!--\; \alpha \; -->}$=0.05 significance level, they should conclude that the proportion has changed from 10%.
d) At the $\mathrm{\alpha <!--\; \alpha \; -->}$=0.05 significance level, they should conclude that the proportion is still 10%.
The correct answer is c but why could it not have been b? Why is it c?

Equations of significance probabilities
Consider a population of independent light bulbs with an exponential lifetime distribution with mean $\mathrm{\mu <!--\; \mu \; -->}$. It is claimed that their expected lifetime is 1000 hours. A definition of a 100(1−$\mathrm{\alpha <!--\; \alpha \; -->}$)% confidence interval obtained from an observation to is the set of all ${\mathrm{\mu <!--\; \mu \; -->}}_0$ which are not rejected in a test of a null hypothesis ${\mathrm{\mu <!--\; \mu \; -->}}_0$ against an alternative hypothesis $\ne <!--\; \ne \; -->$.
One particular light bulb fails after 622 hours. Solve the equations of the two significance probabilities Pr(T "" 622 |${\mathrm{\mu <!--\; \mu \; -->}}_0$) = 0.05 (for a test of ${\mathrm{\mu <!--\; \mu \; -->}}_0$ versus ${\mathrm{\mu <!--\; \mu \; -->}}_0$) and Pr(T "" 622 |${\mathrm{\mu <!--\; \mu \; -->}}_0$ = 0.05 (for a test of ${\mathrm{\mu <!--\; \mu \; -->}}_0$ versus ${\mathrm{\mu <!--\; \mu \; -->}}_0$) for $\mathrm{\mu <!--\; \mu \; -->}$. Determine the range of values of $\mathrm{\mu <!--\; \mu \; -->}$ such that both of the probabilities Pr(T "" 622 | $\mathrm{\mu <!--\; \mu \; -->}$) and Pr(T "" 622 |$\mathrm{\mu <!--\; \mu \; -->}$) are at least 0.05. (This range gives an equi-tailed 90% confidence interval for $\mathrm{\mu <!--\; \mu \; -->}$.)
I don't seem to understand what they mean by 'solve the equations'. Do I have to find a specific value for T or compute Pr(T $\ge <!--\; \ge \; -->$ 622 |${\mathrm{\mu <!--\; \mu \; -->}}_0$), Pr(T $\le <!--\; \le \; -->$ 622 |${\mathrm{\mu <!--\; \mu \; -->}}_0$) and compare with 0.05? I believe I will get the second part after I understand this bit.

Two Tail Hypothesis Test with Variance
test the hypothesis the variance for economists equal the variance for the historians. Use a .05 significance level, a two tail test and the following data :
economist historian
var 120 90
n 46 38