Research Hypothesis

Test of Hypothesis – Large Sample Test Critical Value (Zx) | Level of Significance (x) | | 1% | 5% | Two tailed test | Z = 2.58 | Z = 1.96 | One tailed test | Z = 2.33 | Z = 1.64 |

Q. Z= X- μ σ√n = x- μS.EX

The mean height of a random sample of 100 students is 64” and standard deviation is 3”. Test the statement that the mean height of population is 67” at 5% level of significance.

Solution: We are given n = 100 , X = 64, S = 3 µ = 67 Null HypothesisAlternative Hypothesis H0: μ=67H1: μ ≠67
(Two tailed test) Z= X- μ=64-67 0.3=10
At 50% level of significance, the critical value of Z for two tailed test = 1.96.
Since the calculated value of Z is more that the critical value of Z at 5% level of significance, we reject the null hypothesis and conclude that the population mean cannot be equal 67.

Ex 2: A stenographer claims that she can take dictation at the rate of 120 words per minutes. Can we reject her claim on the basis of 100 trials in which she demonstrate a mean of 116 words with a standard deviation of 15 words (use 5% I.O.S)
Sol.: We are given n = 100 X=116 S=15 μ=120

Null Hypothesis = H0 : µ = 120 (the claim is accepted)
Alternative Hypothesis : H1 : µ < 120 (i.e. the claim is rejected)
S. EX = Sn = 15100 = 1510 =1.5
Z= X- μS. EX = 116-1201.5 = 41.5=2.6 At 5% I.O.S> the critical value of Z for one tailed test = 1.645.
Since calculated of Z > 1.645, we rejected H0 and conclude that the claim of the stenographer is rejected.

Aliter: This question can also be solved by using two tailed test let us have
H0 : µ = 120 and H1 : µ ≠ 120 it is two tailed test.
And H1 : µ ≠ 120 it is two tailed test Z=2.6
At 5% Z .05 = 1.96 (two tailed test)
Since Z> 1.96 we reject H0 and conclude that the claim of stenographer is