# 统计代写|假设检验代写hypothesis testing代考|STAT101

## 统计代写|假设检验代写hypothesis testing代考|Testing one sample mean when the variance is known: P-value

We have studied hypothesis testing for one sample mean when the sample size is large using the critical value (traditional) procedure. The $P$-value procedure for one sample mean using a Z-test will be illustrated employing the same examples presented in Chapter 2, Z-test for one-sample mean, and solved using the critical value procedure. The three situations of hypothesis testing are covered and the $P$-value are calculated.

Example 6.4: The concentration of cadmium of surface water: Example $2.5$ is reproduced “A professor at an environmental section wanted to verify the claim that the mean concentration of cadmium (Cd) of surface water in Juru River is $1.4(\mathrm{mg} / \mathrm{L})$. He selected 35 samples and tested for cadmium concentration. The collected data showed that the mean concentration of cadmium is $1.6$ and the standard deviation of the population is $0.4$. A significance level of $\alpha=0.01$ is chosen to test the claim. Assume that the population is normally distributed.”

The five steps for conducting hypothesis testing employing the $P$-value procedure can be used to test the hypothesis regarding the mean concentration of cadmium of surface water in Juru River. The results of the $P$-value procedure will be compared with the critical value (traditional) procedure.
Step 1: Specify the null and alternative hypotheses
The two hypotheses regarding the mean concentration of cadmium of surface water in Juru River are presented in Eq. (6.4).
$$H_0: \mu=1.4 \text { vs } H_1: \mu \neq 1.4$$
Step 2: Select the significance level $(\alpha)$ for the study
The level of significance is chosen to be $0.01$. The $\mathrm{Z}$ critical values for a twotailed test with $\alpha=0.01$ are $\pm 2.58$.

The two procedures will be used to solve this problem; namely the critical value and $P$-value procedures.
Step 3: Use the sample information to calculate the test statistic value
The test statistic value for the Z-test is used to make a decision regarding the mean concentration of cadmium of surface water in Juru River. The test statistic value using the Z-test formula was calculated to be $2.96$.

## 统计代写|假设检验代写hypothesis testing代考|Computing the P-value for a t-test

This section presents the steps for computing the $P$-value for a t-test including onesided and two-sided tests. The procedure will be illustrated step by step with examples using the general procedure for hypothesis testing.

Example 6.7: Compute the $\boldsymbol{P}$-value to the left of a t value: Compute the $P$-value to the left of a negative $t$ value $(t=-2.02)$ and sample size 16, (left-tailed test). Use a significance level of $0.01(\alpha=0.01)$.

Computing the $P$-value for a left-tailed t-test can be achieved employing the general procedure for testing a hypothesis.
Step 1: Specify the null and alternative hypotheses
The two hypotheses (the null and alternative) for a left-tailed test can be written as presented in Eq. (6.7).
$$H_0: \mu \geq c \text { vs } H_1: \mu<c$$
The hypothesis in Eq. (6.7) represents a one-tailed test because the alternative hypothesis is $H_1: \mu<c$ (left-tailed), where $c$ is a given value.
Step 2: Select the significance level $(\alpha)$ for the study
The level of significance is chosen to be $0.01$.
Step 3: Use the sample information to calculate the test statistic value
The test statistic value for the t-test is given to be $t=-2.02$, otherwise we have to calculate it using a formula.

Step 4: Calculate the $P$-value and identify the critical and noncritical regions for the study

The $P$-value for a t-test can be computed easily by employing the $\mathrm{t}$ table (Table $\mathrm{B}$ in the Appendix) which depends on two values: the degrees of freedom $(d . f)$ and the level of significance $(\alpha)$. The $P$-value for $-2.02$ with $d . f=15$ falls somewhere in the interval $0.025<P$-value $<0.05$ as illustrated below.

We can use the symmetry property of $\mathrm{t}$ distribution to compute the $P$-value to the left of a $t$ value because all values in the table are for the positive value. The position of the required value $(2.02)$ with degrees of freedom $(d . f=15)$ falls between $1.753$ and $2.131$, these two values correspond with the significance level of $0.025$ and $0.05$ for $1.753$ and $2.131$, respectively. Thus, one can say that the $P$-value for $t=-2.02$ and $d . f=15$ is somewhere in the interval $0.025<P$-value $<0.05$. The exact $P$-value $(0.03147346=0.03)$ using a statistical software is presented in Fig. 6.7.

The blue shaded area in Fig. $6.7$ represents the required $P$-value for the value $-2.02$ and the level of significance $(0.01)$ represents the orange shaded area.

# 假设检验代考

## 统计代写|假设检验代写假设检验代考|当方差已知时检验一个样本均值:p值

Example 6.4: The concentration of cadmium of地表水:Example $2.5$ is转载“一位环境学系的教授想要验证Juru河地表水镉(Cd)的平均浓度为$1.4(\mathrm{mg} / \mathrm{L})$的说法。他选取了35个样品进行镉浓度测试。采集的数据显示，镉的平均浓度为$1.6$，人群的标准差为$0.4$。选择显著性水平$\alpha=0.01$来检验该主张。假设总体是正态分布的。

$$H_0: \mu=1.4 \text { vs } H_1: \mu \neq 1.4$$

## 统计代写|假设检验代写假设检验代考|计算t检验的p值

t检验的检验统计值给定为 $t=-2.02$

t检验的$P$ -值可以通过使用$\mathrm{t}$表(附录中的表$\mathrm{B}$)轻松计算出来，它取决于两个值:自由度$(d . f)$和显著性水平$(\alpha)$。$-2.02$和$d . f=15$的$P$ -值位于$0.025<P$ -value $<0.05$的某个区间，如下所示

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