If the sample mean = 16, population mean = 8, sample standard deviation = 4, and sample size = 150, what is the one sided t-score of the sample?

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Answer 1 · 2015-12-10T09:39:45

The Calculated "t' value is $24.5$ $24.5$
Table value (t Value) for 5% level of significance and 149 degrees of freedom is $1.645$ $1.645$

You didn't give the significance level.
Let me use 5% significance level.

Since the sample mean is greater than population mean, we can go for Right Tail test.

Table value (t Value) for 5% level of significance and 149 degrees of freedom is 1.645

Calculate Standard Error

$S E = \frac{s}{\sqrt{n}} = \frac{4}{\sqrt{150}} = \frac{4}{12.25} = 0.3265$ $SE =s/sqrt(n)=4/sqrt(150)=4/12.25=0.3265$

Calculate t value for the given values.

$t = \frac{¯ x - μ}{S E} = \frac{16 - 8}{0.3265} = \frac{8}{0.3265} = 24.5$ $t=(barx-mu)/(SE)=(16-8)/(0.3265)=8/(0.3265)=24.5$

The Calculated "t' value is $24.5$ $24.5$