Two Tailed Test of Hypothesis

Since the population standard deviation is known we will use z table in both cases.
(If it was not known then we would have used t table for sample sizes less than 30 in this case in part a)

a)

Hypothesized Mean=M = 50
Standard deviation =s= 20
sample size=n= 24
sx=standard error of mean=s/square root of n= 4.0825 = ( 20 /square root of 24)
Significance level=alpha (a) = 0.05
No of tails= 2
This is a 2 tailed test because we are testing M 1not equal to M 2

sample mean= 56.53
Z at the 0.05 level of significance 2 tailed test = 1.96

Upper acceptance limit= M +z*sx= 58.0017 =50+1.96*4.0825
Lower acceptance limit= M -z*sx= 41.998 =50-1.96*4.0825

Accept Null Hypothesis since the sample mean is within acceptance region

z=(sample mean-M )/sx= 1.5995 =(56.53-50)/4.0825
Prob-value corresponding to z= 1.5995 is 5.49%
Therefore area in both the tails= 10.98%
P value= 10.98% or 0.1098 ( this is more than the significance level of 0.05 )

b
Hypothesized Mean=M = 50
Standard deviation =s= 20
sample size=n= 36
sx=standard error of mean=s/square root of n= 3.3333 = ( 20 /square root of 36)
Significance level=alpha (a) = 0.05
No of tails= 2
This is a 2 tailed test because we are testing M 1not equal to M 2

sample mean= 56.53
Z at the 0.05 level of significance 2 tailed test = 1.96

Upper acceptance limit= M +z*sx= 56.5333 =50+1.96*3.3333
Lower acceptance limit= M -z*sx= 43.467 =50-1.96*3.3333

Accept Null Hypothesis since the sample mean is within acceptance region

z=(sample mean-M )/sx= 1.959 =(56.53-50)/3.3333
Prob-value corresponding to z= 1.959 is 2.51%
Therefore area in both the tails= 5.02%
P value= 5.02% or 0.0502 ( this is more than the significance level of 0.05).

See attached file.

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