A random sample of 8 observations taken from a population that is normally distributed produced a sample mean of 44.98 and a standard deviation of 6.77. Find the critical and observed values of t and...

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Test: µ 6= 50 with t distn
Hypothesis test:
H0 : µ = 50 (Null Hypothesis)
Ha : µ 6= 50 (Alternative Hypothesis, also called H1)
This is two tailed test.
x = 44.98
s = 6.77
n = 8
Significance Level, α = 0.05
Degrees of freedom df = n− 1 = 7
Test statistic, tobserved =
x− µ
s/
√
n
= 44.98− 50
6.77/
√
8
= −2.097
Test statistic = −2.097
Using t-table, we can observe that the p-value is between 0.05 and 0.10
Using excel we find p − value = P (t < | − 2.097|) = 2*t.dist(2.097, 7,1 ) =
0.0741665608
tcritical = tα/2,df = t0.025,df=7 = 2.365 (from t-table, two-tails, d.f . = 7 )
Critical Value, tc = ± 2.365
Rejection criteria: Rejection region is in direction of Alternative...