Question

The values of two random samples are given below.
Sample I: 15 25 16 20 22 24 21 17 19 23
Sample II: 35 31 25 38 26 29 32 34 33 27 29 31
Can we conclude that the two samples are drawn from the same population? Test at 5% level of significance
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Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Samples are not from the same population.

Step-by-step explanation:

We shall use both the F-test and t-test to draw a conclusion.We have $n_1 = 12$,

$x_1 = 20.2$,${s_1}^{2} = 10.56$

${\sigma_1}^{2} = \frac{n_1 {s_1}^{2} }{n_1 - 1} = 11.7333$

$n_2 = 12$,$x_2 = 30.8333$,${s_2}^{2} = 13.6409$,${\sigma_2}^{2} = \frac{n_2 {s_2}^{2} }{n_2 - 1} = 14.8810$

Note that ${\sigma_2}^{2} > {\sigma_1}^{2}$

We define,Null hypothesis $H_0 :{\sigma_1}^{2} = {\sigma_2}^{2}$

Alternate hypothesis

$H_1 :{\sigma_1}^{2} ≠ {\sigma_2}^{2}$

At 5% level of significance

$F_{0.05}(11 \: 9) = 3.11$

The F-statistic is given by

$F = \frac{ {\sigma_2}^{2} }{ {\sigma_1}^{2} } = \frac{14.8810}{11.7333} = 1.268 < 3.11$

We accept the null hypothesis.We conclude that the difference between estimates of population variances is not significant.

Now,we use the t-test.Define $H_0: \: \mu _1=\mu_2$and $H_1: \: \mu _1≠\mu_2$

The estimate of the population variance is given by.

${\sigma}^{2} = \frac{n_1 {s_1}^{2} + n_2 {s_2}^{2} }{n_1 + n_2 - 2} = \frac{10 \times (10.56) + 12 \times (13.6409)}{20} = 13.4645$

The t-statistic is given by$t = \frac{x_1 - x_2}{ \sigma\sqrt{ \frac{1}{ \sqrt{n_1} } + \frac{1}{n_2} } } = - 6.67$With $v=n_1+n_2-2=20$

degrees of freedom, At 5% level of significance,we get from given table.The value of t as Since $|t| = 6.76 > 2.086$,we reject the null hypothesis.The means of the two samples and hence the means of the two populations differ significantly. Sample are not from the same population.