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

F-Test

Fisher is Z Test

T-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

t=0.525

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

= 14.52

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...