One-Sample Z-Test for Mean

189. One-Sample Z-Test for Mean

Warning: Pipulation standard deviation, $𝜎$ , is known.

\begin{aligned} 𝐻_{0} & : 𝜇 \\ 𝐻_{𝑎} & : 𝜇 \end{aligned}

Example

Your favorite candy company claims their chocolate bars have a mean weight of 200 grams, and their population standard deviation is 5 grams. You’ve noticed that they’ve seemed lighter than usual. You take a simple random sample of 35 candy bars and find they have a mean weight of 198 grams. At $𝛼$ = 0.05 test the company’s claim.

\begin{aligned} 𝐻_{0} & : 𝜇 \geq 200 \\ 𝐻_{𝑎} & : 𝜇 < 200 \end{aligned}

𝑧 = \frac{\bar{𝑥} - 𝜇}{𝜎 / \sqrt{𝑛}}

Where:

$\bar{𝑥} = 198$
$𝜇 = 200$
$𝜎 = 5$
$𝑛 = 35$
$𝜎 / \sqrt{𝑛} = 5 / \sqrt{35} = 0.8452$

𝑧 = \frac{198 - 200}{5 / \sqrt{35}} = −2.3663

𝑃 (𝑧 \leq −2.3663) = 0.009

\begin{aligned} 𝑧 -value & < critical-value \\ −2.3663 & < −1.6449 \end{aligned}

\begin{aligned} 𝑝 -value & < 𝛼 -value \\ 0.009 & < 0.05 \end{aligned}

Reject $𝐻_{0}$ . At $𝛼 = 0.05$ there is sufficient evidence that the mean weight of the chocolate bars is less than 200 grams.

Code

  import numpy as np
  from scipy.stats import norm

  mu = 200.0
  sigma = 5.0
  xbar = 198
  n = 35
  alpha = 0.05

  cv = norm.ppf(alpha)

  z = (xbar - mu) / (sigma / np.sqrt(n))
  p = norm.cdf(z)

  if z <= cv:
      print("Reject null hypothesis")
  else:
      print("Fail to reject null hypothesis")