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    Two Tailed Test of Hypothesis

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    In a normal population with sigma=20, a random sample of size n gave X(mean)=56.53. At the 5% significance level, make a two-tailed test of the null hypothesis mu=50, when (a) n=24 (b) n=36.

    In each case report the P value.

    © BrainMass Inc. brainmass.com December 24, 2021, 5:06 pm ad1c9bdddf
    https://brainmass.com/statistics/hypothesis-testing/tailed-test-hypothesis-28375

    SOLUTION This solution is FREE courtesy of BrainMass!

    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.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 5:06 pm ad1c9bdddf>
    https://brainmass.com/statistics/hypothesis-testing/tailed-test-hypothesis-28375

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