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 rejected.
Test of Hypothesis – Small Sample Tests
n < 30
Small Sample Test
Fisher is Z Test
Student’s Test Variance Ratio Test
T – Test
T = X- μS n
= √(X- X)2n-1 when deviations are taken from actual mean. Degree of freedom r=n-1
Q1: A group of 5 patients treated with medicine a weight : 42, 39, 48, 60 and 41 kg. in a light of the above data, discuss the suggestion that mean weight of the population is 48 kg. test at 5% level of significance. (given the table value of t for 4 d.f. at 5% level is 2.776) Sol.:
Weight (X)| X=46, (X- X| (X - X)2|
42| − 4| 16|
39| − 7| 49|
48| + 2| 4|
60| + 14| 196|
41| − 5| 25|
Σx=230, n=5| | Σ (X- X)2 = 290|
X= ΣXn= 2305=46
S = Σ(X- X)2n-1 = = 2905-1 = 8.514
Applying t – test
t = X- μS n
t=46-488.514 5 = 2 X 2.2368.514
Degree of Freedom - r=n-1 = 5 – 1 = 4
For r = 4 t = 0.05 for two tailed test = 2.776
Since, the calculated value of t is less than the table value, we accept the null hypothesis and therefore conclude that the mean weight in the population is 48 kg. Q2.: Sixteen oil tins are taken at random from an automatic filling machine. The mean weight of the tins is 14.5 kg with a standard deviations of 0.40 kg. Does the sample mean differ significantly from the intended weight of 16 kg? Sol.: we are given
N = 16, X=14.5 S=0.40 Kg.
S = nn-1 S2 = 1616-1 X (0.4)2
When sample standard deviation
= 2.5616 = 0.413
S = Standard deviation taken from sample
S = Standard deviations are taken from actual mean.
S = Σ(X- X)2n-1
Applying t – test
t = X- μS n = 14.5-160.413 16
Degree of freedom r=n-1 = 16 – 1 = 5
For r = 15 t = 0.05 for two tailed test = 2.131.
Since, the calculated value of t = 14.52.
So calculated value t = 14.52 > table of t. we reject...