A sample of 25 has mean of 775 and standard deviation of 230, using 2% level, fo

Question

A sample of 25 has mean of 775 and standard deviation of 230, using 2% level, for testing the claim that population mean is different from 850, find:
1) Test Type And Critical Value 2) Observe Value and Decision for making confidence interval of population mean 3) Marginal Error 4) The Interval A sample of 25 has mean of 775 and standard deviation of 230, using 2% level, for testing the claim that population mean is different from 850, find:
1) Test Type And Critical Value 2) Observe Value and Decision for making confidence interval of population mean 3) Marginal Error 4) The Interval
1) Test Type And Critical Value 2) Observe Value and Decision for making confidence interval of population mean 3) Marginal Error 4) The Interval

Explanation / Answer

Solution1:

1) Test Type And Critical Value

Hypothesis tets for singlemean

alpha=1-0.98=0.02

alpha2=0.02/2=0.01

df=n-1=25-1=24

t critical for 24 df and 0.01 level of significance

=2.492

Solution2:

observed value=t

t=sample mean-popualtion mean/sample sd/sqrt(n)

=775-850/230/sqrt(25)

=-1.630

alpha=0.02

p=0.1161

p>0.02

Fail to reject Null hypothesis .

Accept null hypothesis

There is no sufficient statistical evidence at 2% level of signiifcance to conclude that  population mean is different from 850,

USing Ti 83 cal

98% confidence interval for population mean is

660.36 and 889.64

Solution3:

marginal error=t crit*samplesd/sqrt(n)

=2.492(230/sqrt(25)

=114.632

Solution4:

the interval is

lower limit=sample mean-Marginal Error=775-114.632=660.368

and

upper limit=775+114.632=889.632

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