The conclusion that can be drawn from the population is that it's an unusual event.
From the information, the normal population with unknown variance has a mean of 20. is one likely to obtain a random sample of size 9 from this population with a mean of 24 and a standard deviation of 4.1.
The test statistic will be:
t = (Population mean - Sample mean) / Deviation / âś“number
t = (24 - 20) / 4.1 / âś“9.
t = 2.93
Degree of freedom = n - 1
= 9 - 1
= 8
The p value is 0.0095 from a distribution table.
Since the p value is less, it's an unusual event.
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