Hypothesis testing for single population mean when the variance \(\sigma^{2}\) is known¶

Assume that we are interested in making an inference about the mean of a normally distributed single population with known variance and we have a random sample of size n drawn from this population:

\[ X_{1}, X_{2}, \dots, X_{n} \sim N(\mu, \sigma^{2}). \]

When the sample is coming from a normally distributed population with known variance \(\sigma^{2}\), under \(H_{0}\), the test concerning the population mean \(\mu\), is based on the following test statistic:

\[ \begin{aligned} Z^{*} &=\frac{\bar{X}-\mu_{0}}{\sigma/\sqrt{n}}\sim N(0,1) \end{aligned} \]

where \(\bar{X}\) is the sample mean, \(\mu_{0}\) is the hypothesized population mean value, and \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
For a two-sided test:

\[\begin{split} \begin{aligned} H_{0} &: \mu = \mu_{0} \nonumber \\ H_{1} &: \mu \neq \mu_{0} \nonumber \end{aligned} \end{split}\]

Reject \(H_{0}\) if \(z^{*} \leq -z(1-\alpha/2)\) OR if \(z^{*} \geq z(1-\alpha/2)\), otherwise fail to reject \(H_{0}\).
Here, \(z(1-\alpha/2)\) refers to the upper \(1-\alpha/2\) percentage point of the standard normal distribution.
For a one-sided upper-tail test:

\[\begin{split} \begin{aligned} H_{0} &: \mu \leq \mu_{0} \nonumber \\ H_{1} &: \mu > \mu_{0} \nonumber \end{aligned} \end{split}\]

Reject \(H_{0}\) if \(z^{*} \geq z(1-\alpha)\), otherwise fail to reject \(H_{0}\).
Here, \(z(1-\alpha)\) refers to the upper \(\alpha\) percentage point of the standard normal distribution.
For a one-sided lower-tail test:

\[\begin{split} \begin{aligned} H_{0} &: \mu \geq \mu_{0} \nonumber \\ H_{1} &: \mu < \mu_{0} \nonumber \end{aligned} \end{split}\]

Reject \(H_{0}\) if \(z^{*} \leq z(\alpha)=-z(1-\alpha)\), otherwise fail to reject \(H_{0}\).
The P-value is the probability that we would observe a more extreme statistic than we did if the null hypothesis were true.
If \(z^{*}\) is the computed value of the test statistic, the P-value is:

\[\begin{split} P-value = \begin{cases} 2[1-\Phi(|z^{*}|)] & \quad {for} \quad H_{0} : \mu = \mu_{0} \quad \text{vs} \quad H_{1} : \mu \neq \mu_{0} \nonumber \\ 1-\Phi(z^{*}) & \quad {for} \quad H_{0} : \mu \leq \mu_{0} \quad \text{vs} \quad H_{1} : \mu > \mu_{0} \nonumber \\ \Phi(z^{*}) & \quad {for} \quad H_{0} : \mu \geq \mu_{0} \quad \text{vs} \quad H_{1} : \mu < \mu_{0} \nonumber \end{cases} \end{split}\]

where \(\Phi (z^{*}) = Pr(Z \leq z^{*})\) is the standard normal cumulative distribution function.
Decision rule: If \(P-value\) \leq \(\alpha\), then reject \(H_{0}\), otherwise fail to reject \(H_{0}\).

Example¶

At a certain production facility that assembles computer keyboards, from past experience the assembly time is known to follow a normal distribution with mean (\(\mu\)) of 130 seconds and standard deviation (\(\sigma\)) of 15 seconds.
The new production supervisor suspects that the average time to assemble the keyboards does not quite follow the specified value.
To examine this problem, he measures the times for 100 assemblies and found that the sample assembly time average (\(\bar{X}\)) is \(126.8\) seconds.
Can the supervisor conclude at the \(5\%\) level of significance that the mean assembly time of 130 seconds is incorrect?

# define the hypothesis testing function for testing single normal population
# mean when pop. variance is known

import numpy as np
import scipy.stats as stats
    
def OneSampZ(Xbar, sigma1, n1, mu0 = 0, alpha = 0.05, direction = "two_sided"):
    
    z_star = (Xbar-mu0)/np.sqrt(sigma1**2/n1)

    if (direction == "two_sided"): 
        p_val = 2*(1 - stats.norm.cdf(abs(z_star)))
    elif (direction == "one_sided_upper_tail"): 
        p_val = (1 - stats.norm.cdf(z_star))
    else:
        p_val = stats.norm.cdf(z_star)
    
    if (p_val < alpha):
        Hypothesis_Status = 'Reject Null Hypothesis : Significant' 
    else:
        Hypothesis_Status = 'Do not reject Null Hypothesis : Not Significant'
 
    print (Hypothesis_Status)

    z_star = np.round(z_star,4)
    p_val = np.round(p_val,4)
    
    return z_star, p_val

OneSampZ(Xbar = 126.8, sigma1 = 15, n1 = 100, mu0 = 130, alpha = 0.05, direction = "two_sided")

Reject Null Hypothesis : Significant

(-2.1333, 0.0329)

Spring 2022

Hypothesis testing for single population mean when the variance \sigma^{2} is known

Contents

Hypothesis testing for single population mean when the variance \(\sigma^{2}\) is known¶

Example¶

Hypothesis testing for single population mean when the variance \(\sigma^{2}\) is unknown¶

Example¶

Example¶

Session Info¶