Management Looks magazine did a survey of midmanagement people to see if there i

Question

Management Looks magazine did a survey of midmanagement people to see if there is a difference in the proportion of women and men going to a hair stylist regularly. A random sample of 48 midmanagement men showed that 33 went to stylists regularly, while a random sample of 41 midmanagement women showed that 25 went to stylists regularly. Test the claim that there is no difference in the proportion of midmanagement men and women who go to hair stylists regularly. Use a 0.05 level of significance. a. List the null and alternate hypotheses in terms of the appropriate population parameter using =, <, or >. b. Find the critical value(s) which determine(s) the rejection and acceptance regions. c. Compute the relevant test statistic and the P-value. d. Draw a conclusion. Do you reject the null hypothesis or not?

Explanation / Answer

Solution

Let X = mid-management men out of a sample of 48 who went to stylists regularly

and Y = mid-management women out of a sample of 41 who went to stylists regularly

Then, X ~ B(n1, p1), and Y ~ B(n2, p2) where n1 and n2 are sample sizes and p1 = probability of a mid-management man going to stylists regularly and p2 = probability of a mid-management woman going to stylists regularly, which are also equal to the proportion of a mid-management men and women going to stylists regularly in the population.

Claim :

There is no difference in the proportion of mid-management men and women who go to hair stylists regularly.

Hypotheses:

Null H0 : p1 = p2   Vs HA : p1 p2

Test Statistic:

Z = (p1cap – p2cap)/[pcap(1 - pcap){(1/n1) + (1/n2)}] where p1cap and p2cap are sample proportions, n1, n2 are sample sizes and pcap = {(n1 x p1cap) + (n2 x p2cap)}/(n1 + n2).

Calculations:

p1cap = 33/48 = 0.6875

p2cap = 25/41 = 0.6098

pcap = (33 + 25)/(48 + 41) = 58/89 = 0.6517

Zcal = (0.6875 - 0.6098)/[0.6517 x 0.3483{(1/48) + (1/41)}]

= 0.0777/(0.22698711 x 0.0452)

= 0.0777/0.0102598174

= 0.0777/0.1013

= 0.7671.

Distribution, Critical Value and p-value:

Under H0, distribution of Z can be approximated by Standard Normal Distribution

So, given a level of significance of %, Critical Value = upper (/2)% of N(0, 1), and

p-value = P(Z > | Zcal |)

Using Excel Functions of N(0, 1) for = 5%,

Critical Value = 1.96 and

p-value = P(Z > | 0.7671 |) = 0.443

Decision Criterion (Rejection Region):

Reject H0, if | Zcal | > Zcrit or if p-value < .

Decision:

Since | Zcal | < Zcrit, p-value > , H0 is accepted.

Conclusion :

There is enough evidence to suggest that the claim is valid. i.e., there is no difference in the proportion of mid-management men and women who go to hair stylists regularly.

DONE

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