Question

10. A normal population with an 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? If not what conclusion would you draw. Here are some integrals you may find helpful, where h(t) is the probability distribution function for a t-distribution with 8 degrees of freedom h(t)dt = ht) dt = h(t) dt 0.919 0.653 0.983

Answer 1

Solution HHpotesis be Ho Ha: Sinee, popula kon sAd unknown aud dev test populaton used. Here saple si e So t teat meau should be So we ttasaupling da laibuhen normal Cau agume vmeau Here we have fllooing in tos S=4. test stat So 6i 1l be 24-20 2.93 Hene test ight tai led and tte test PR- 9-1-8. So the Saut pla 8reaber meau ttau a ual to 24 .O095 Smce p-value s less so it is au urual eveng. Tha t ot Ghely to obtain 9audom saupla size a