Monna and co-workers used radioactive isotopes to date sediments from lakes and estuaries. To verify this method they analyzed a 208Po standard known to have an activity of 77.5 decays/min, obtaining the following results 77.09, 75.37, 72.42, 76.84, 77.84, 76.69, 78.03, 74.96, 77.54, 76.09, 81.12, 75.75

Determine whether there is a significant difference between the mean and the expected value at αâ=â0.05.

No significant difference between mean and Expected value

Step-by-step explanation:

Hypothesis

H0 : u = 77.5

H1: u is not equal to 77.5

Alpha = 0.05

Mean = Σxi/n

= 77.09+75.37+72.42+76.84+77.84, +76.69+78.03+74.96+77.54+76.09+81.12+75.75/12

= 919.74/12

= 76.645

We get the variance s² = xi² -n(barx)²/n-1

= 77.09²+75.37²...75.75²-12(76.645)²/11

When we solve this out

We get 4.3486818182

T test = barx - u/(s/√n)

= (76.645-77.5)/√4.3486818182/12

= -1.420293

T critical = Tn-1, alpha/2

= 2.200985

The test statistic is less than 2.201 so we accept the null hypothesis at 0.05 level of significance. And then conclude that there is No significant difference between mean and Expected value.

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