# 统计代写|假设检验代写hypothesis testing代考|MATH214

## 统计代写|假设检验代写hypothesis testing代考|Confidence interval for one population proportion

Hypothesis testing for one population proportion was studied in Chapter 4, Z-test for one sample proportion. Calculating the confidence interval for one population proportion can be achieved by employing the general procedure for calculating the confidence interval presented earlier.

Consider a random sample that is selected from a normally distributed population and $p$ represents the proportion of a characteristic in the population. A confidence interval regarding the population proportion can be calculated using binomial distribution. When a large sample size is used, then binomial distribution can be approximated to normal distribution; the probability of success is small and $n p>5$ and $n q \geq 5$. The mathematical formula for computing the confidence interval for one population proportion is presented in Eq. (7.5).
$$\hat{p} \pm Z_{\frac{\beta}{2}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$

The confidence interval can be written using another form:
$$\hat{p}-Z_{\frac{\hat{\beta}}{2}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \leq p \leq \hat{p}+Z_{\frac{\hat{\beta}}{2}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$
where
$\hat{p}$ represents the sample proportion;
$p$ represents the population proportion;
$Z_{\frac{a}{2}}$ represents the $Z$ critical value;
$n$ represents the sample size; and
$Z_{\frac{a}{2}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ represents the margin of error $(E)$.

• The margin of error $(E)$ can be used to write the confidence interval $\hat{p}-E \leq p \leq p+E$.
• The lower one-tailed confidence interval is
$$\hat{p}-Z_\alpha \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \leq \mu$$
• The upper one-tailed confidence interval is
$$\mu \leq \hat{p}+Z_\alpha \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$

## 统计代写|假设检验代写hypothesis testing代考|Confidence interval for one population variance

Hypothesis testing for one population variance was studied in Chapter 5, Chisquare test for one sample variance. Calculating the confidence interval for one population variance can be achieved by employing the general procedure for calculating the confidence interval presented earlier.

Consider a random sample that is selected from a normally distributed population and $\sigma^2$ represents the variance of the population. A confidence interval regarding the population variance can be calculated using chi-square distribution. The mathematical formula for computing the confidence interval for one population variance is presented in Eq. (7.6).
$$\frac{(n-1) S^2}{\chi_R^2} \leq \sigma^2 \leq \frac{(n-1) S^2}{\chi_L^2}$$
Follow a chi-square distribution with $d . f=(n-1)$.
where, $\chi_R^2=\chi_{\left(\frac{a}{2}, n-1\right)}^2$ which represents the right tail and $\chi_L^2=\chi_{\left(1-\frac{a}{2}, n-1\right)}^2$ represents the left tail.
The confidence interval for the standard deviation is:
$$\sqrt{\frac{(n-1) S^2}{\chi_R^2}} \leq \sigma \leq \sqrt{\frac{(n-1) S^2}{\chi_L^2}}$$

• The lower one-tailed confidence interval for the population variance is $\frac{(n-1) 5^2}{x_R^2} \leq \sigma^2$
• The upper one-tailed confidence interval for the population variance is $\sigma^2 \frac{x_R^2}{\leq} \frac{(n-1), 5^2}{x_L^2}$
Example 7.7: The wind speed: A professor at a research center wants to know the range of variance of wind speed at the place where she works. She selected 20 days and the wind speed was measured. The results showed that the variance of the wind speed is $0.27\left(\mathrm{~m} \mathrm{~s}^{-1}\right)$. Calculate the confidence interval for the variance of wind speed using $0.01$ as the level of significance. Assume that the population is normally distributed.

The general procedure for calculating the confidence interval with the five steps will be used to calculate the confidence interval for the variance of wind speed.

# 假设检验代考

## 统计代写|假设检验代写hypothesis testing代考|Confidence interval for one population proportion

$$\hat{p} \pm Z_{\frac{1}{2}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$

$$\hat{p}-Z_{\frac{\beta}{2}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \leq p \leq \hat{p}+Z_{\frac{\beta}{2}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$

$\hat{p}$ 代表样本比例;
$p$ 代表人口比例;
$Z_{\frac{a}{2}}$ 代表 $Z$ 临界值;
$n$ 代表样本量; 和
$Z_{\frac{a}{2}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ 表示误差范围 $(E)$.

• 误差幅度 $(E)$ 可以用来写置信区间 $\hat{p}-E \leq p \leq p+E$.
• 下单尾置信区间为
$$\hat{p}-Z_\alpha \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \leq \mu$$
• 上单尾置信区间为
$$\mu \leq \hat{p}+Z_\alpha \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$

## 统计代写|假设检验代写hypothesis testing代考|Confidence interval for one population variance

$$\frac{(n-1) S^2}{\chi_R^2} \leq \sigma^2 \leq \frac{(n-1) S^2}{\chi_L^2}$$
䒼循卡方分布 $d \cdot f=(n-1)$.

$$\sqrt{\frac{(n-1) S^2}{\chi_R^2}} \leq \sigma \leq \sqrt{\frac{(n-1) S^2}{\chi_L^2}}$$

• 总体方差的下单尾置信区间为 $\frac{(n-1) 5^2}{x_R^2} \leq \sigma^2$
• 总体方差的上单尾置信区间为 $\sigma^2 \frac{x_R^2}{\leq} \frac{(n-1), 5^2}{x_L^2}$
例 7.7: 风速: 某研究中心的教授想知道她工作地点的风速变化范围。她选择了20天，测量了风速。 结果表明，风速的方差为 $0.27\left(\mathrm{~m} \mathrm{~s}^{-1}\right)$. 使用计算风速方差的置信区间 $0.01$ 作为显着性水平。假设总 体呈正态分布。
用五步计算置信区间的一般程序将用于计算风速方差的置信区间。

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