## 统计代写|假设检验代写hypothesis testing代考|Compute the P-value for a chi-square test

We have studied hypothesis testing for one sample variance (standard deviation) using the critical value (traditional) procedure. The $P$-value procedure for one sample variance or standard deviation using a chi-square test will be illustrated employing examples to cover various situations of hypothesis testing.

Example 6.16: Compute the $P$-value to the left of a $\chi^2$ value: Compute the $P$-value to the left of a chi-square value $\left(\chi^2=3.92\right)$ and sample size 12 (left-tailed test). Use a significance level of $0.01(\alpha=0.01)$.

Computing the $P$-value for a left-tailed chi-square 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 left-tailed test can be written as presented in (6.16).
$$H_0: \sigma^2 \geq c \text { vs } H_1: \sigma^2<c$$
The hypothesis in Eq. (6.16) 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 of $\chi^2$ is given to be $\chi^2=3.92$, 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 chi-square test can be computed easily by employing the chisquare table (Table $\mathrm{C}$ in the Appendix) which depends on two values: the degrees of freedom $(d . f)$ and the level of significance $(\alpha)$. The $\chi^2$ critical value for the lefttailed is $\chi_{(1-\alpha, d, f)}^2=\chi_{(1-0.01,11)}^2=3.053$. The $P$-value for $3.92$ with $d . f=11$ falls somewhere in the interval $0.025<P$-value $<0.05$ as illustrated below.

## 统计代写|假设检验代写hypothesis testing代考|Testing one-sample population variance

We have studied hypothesis testing for one sample variance (standard deviation) using the critical value (traditional) procedure. The $P$-value procedure for one sample variance (standard deviation) using a chi-square test will be illustrated employing the same examples presented in Chapter 5, Chi -square test for one sample variance, and solved using the critical value procedure. The three situations of hypothesis testing are covered, and the $P$-values are calculated.

Example 6.19: The concentration of total suspended solids of surface water: Example $5.4$ is reproduced “A researcher at an environmental section wishes to verify the claim that the variance of total suspended solids concentration (TSS) of Beris dam surface water is $1.25(\mathrm{mg} / \mathrm{L})$. Twelve samples were selected and the total suspended solids concentration was measured. The collected data showed that the standard deviation of total suspended solids concentration is $1.80$. 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 variance of total suspended solids concentration in the surface water of Beris dam. 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 variance of total suspended solid concentration in the surface water of Beris dam are presented in Eq. (6.19).
$$H_0: \sigma^2=1.25 \text { vs } H_1: \sigma^2=1.25$$
Step 2: Select the significance level $(\alpha)$ for the study
The level of significance is chosen to be $0.01$. The $\chi^2$ critical values for a twotailed test with $\alpha=0.01$ are:

The left-tailed value: The chi-square critical value for a left-tailed test for d. $f=11$ and $1-\frac{\alpha}{2}=0.995$ is $2.603, \chi_{\left(1-\frac{\alpha}{2}, d f\right)}^2=\chi_{(0.995,11)}^2=2.603$

The right-tailed value: The chi-square critical value for a right-tailed test for d. $f=11$ and $\frac{\alpha}{2}=0.005$ is $26.757, \chi_{\left(\frac{a}{2} d, f\right)}^2=\chi_{(0.005,11)}^2=26.757$

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 $\chi^2$-test is used to make a decision regarding the variance of total suspended solids concentration in the surface water of Beris dam. The test statistic value using the chi-square formula was calculated to be $=28.512$.

# 假设检验代考

## 统计代写|假设检验代写hypothesis testing代考|Compute the P-value for a chi-square test

$$H_0: \sigma^2 \geq c \text { vs } H_1: \sigma^2<c$$

$(\alpha)$. 这 $\chi^2$ 左尾的临界值为 $\chi_{(1-\alpha, d, f)}^2=\chi_{(1-0.01,11)}^2=3.053$. 这 $P$-价值 $3.92$ 和 $d . f=11$ 渃在区间的某 处 $0.025<P$-价值 $<0.05$ 如下图所示。

## 统计代写|假设检验代写hypothesis testing代考|Testing one-sample population variance

$$H_0: \sigma^2=1.25 \text { vs } H_1: \sigma^2=1.25$$

$$2.603, \chi_{\left(1-\frac{\alpha}{2}, d f\right)}^2=\chi_{(0.995,11)}^2=2.603$$

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