20250505 7.3 One-Sample Test for the Mean

All data are from Statistic with Technology by Dr. Kozak

Example  7.3.1 test of the mean using the formula 

158180150137109
225122138145180
118118126140165
150170105154118

mean 145.4
sample SD = (= STDEV.S) 29.27438255

(145 - 100) ÷ (29.27 ÷ √(20)) ≈ 6.937

t-table we can use:

Take note:  

You can reject a null hypothesis when a p-value is less than or equal to your significance level.
If our test statistic is:
positive and greater than the critical value, then we have sufficient evidence to reject the null hypothesis and accept the alternative hypothesis.
positive and lower than or equal to the critical value, we must accept the null hypothesis.

Information from:

HOMEWORK 1
The Kyoto Protocol was signed in 1997, and required countries to start reducing their carbon emissions. The protocol became enforceable in February 2005. In 2004, the mean CO2 emission was 4.87 metric tons per capita. Table  7.3.3
  contains a random sample of CO2 emissions in 2010 ("CO2 emissions," 2013). Is there enough evidence to show that the mean CO2 emission is lower in 2010 than in 2004? Test at the 1% level.

Ho: co2_2010=co2_2004
HA: CO2_2010>CO2_2004
a=0.01

÷ (2.97 ÷ √(25)) ≈ -2.39057239057

We can use this tool for calculation: 

You can visit I-Yuan Chiang Youtube Channel for more information

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