If you are using z-distribution for a 1-tail statistical test at the usual 0.05

Question

If you are using z-distribution for a 1-tail statistical test at the usual 0.05 (= 5%) level of significance, the number z_cak you compute is Z_calc = 2.12. What is your conclusion for the corresponding test? Explain! If you are using a t-distribution with df = 11 for a 2-tail statistical test, and the number t_calc you compute is t_calc = 2.10. What is your conclusion for die corresponding test? Show your logic. A psychologist is studying the impact of divorce on the social adjustment of teenagers. Social adjustment ratings are shown below for teenagers whose parents are divorced, and for those whose parents are still married. If you assign low ranks to low scores, what is the sum of ranks you would use to test the significance of Mann-Whitney test? In each of the pictures below the shaded area corresponds to a probability. P(z > z_cak) = 0.45 matches picture

Explanation / Answer

a) Zcal = 2.12

alpha = 0.05

Zcal is positive implies that p-value is 1 - NORMSDIST(z). (In EXCEL)

where z is the test statistic value.

p-value = 0.02

This is right tailed test.

hypothesis is,

H0 : mu = 0 Vs H1 : mu > 0

Reject H0 at 5% level of significance.

Conclusion : Population mean is greator than 0.

b) t-distribution has df = 11 has 2-tailed statistical test has tcal = 2.10

Hypothesis is,

H0 : mu = 0 Vs H1 : mu 0

By using P-value we conlude the test.

EXCEL syntax :

=TDIST(x, deg_freedom, tails)

x is the test statistic value.

deg_freedom = 11

tails = 2

P-value = 0.06

If we assume alpha is 5%.

P-value > 0.05

At 5% level of significance accept H0.

c) Mann whitney test :

Give ranks to divorced parents and married parents.

3.5

The hypothesis for the test is,

H0: the distribution of scores for the two groups are equal

HA: the distribution of scores for the two groups are not equal

Sum of ranks for divoeced parents = 4.5 + 6.5 + 4.5 + 8 + 3 + 2 + 6.5 + 1 = 36

Sum of ranks for married parents = 12 + 2 + 5.5 + 5.5 + 10.5 + 1 + 8 + 3.5 + 8 + 10.5 + 8 + 3.5 = 78

Here larger rank is rank sum for married parents that is 78.

n1 = number of divorced parents = 8

n2 = number of married parents = 12

The test statistic is,

U = n1*n2 + n2*(n2+1)/2

U = 8*12 + (12*13)/2

U = 96 - 78 = 18

d) P(Z > Zcal) = 0.45

Here we have given probability from that we can calculate Zcal by using EXCEL :

=NORMSINV(probability)

where probability = 0.45

Zcal = -0.1257

This Z value is on left side of the normal curve.

So option b) is correct.

divorced ranks married ranks 4 4.5 9 12 5 6.5 4 2 4 4.5 6 5.5 6 8 6 5.5 3 3 8 10.5 2 2 3 1 5 6.5 7 8 1 1 5 3.5 7 8.000 8 10.5 7 8 5

3.5

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