Question

Problem Solving 2. (Finding a job Through Networking in a survey of randomly selected 40 workers, to the question what is the strategy for finding jobs after graduation?", their answers are given as follows: N Networking R-Regular application A Advertisement 0 Other means INNAR NNN ON ANNNNRO N | R | N | NTN | N | NN | R I R NINH | ANIN | N (a) What is the underlying population of interest? What is the variable? Is the variable qualitative or quantitative? What is the highest level of measurement can be used for this data? (b) Construct a frequency table including relative frequencies in percent (%). What strategy is most useful? (c) Use the sample data with a q 0.03 significance level to test the claim that most (more than 50%) workers get their jobs through networking recently. What does the result suggest about the strategy for find a job after graduation? Hint: Run a hypothesis test using the steps: Step 1: Write the null and alternative hypotheses in symbolic form. Step 2: Check the requirements Step 3. Determine the mean and standard deviation of the sampling distribution of the sample statistic what test statistie should be used, assuming the null hypothesis is true Step 4. Determine the type of test, the critical value and sketch the rejection region Step 5. Calculate the sample statistic value.

Answer 1

Solution :

(a)

(i) The underlying population of interest: All persons who are finding jobs after education

(ii) The variable is: trategy for finding jobs after education

(iii) The variable is : qualitative

(iii) Highest level of measurement: Nominal

(b)

(i)

Frequency Table giving relative frequencies in percent (%) is as follows:

Strategy	Frequency	Relative frequencies in percent (%)
N - Networking	27	67.5
R - Regular Application	6	15.0
A - Advertisement	4	10.0
O - Other means	3	7.50
Total	40	100.00

(ii)

Strategy most useful: N - Networking

(c)

Step 1:

H0: Null Hypothesis: p $\leq$ 0.5 (Not more tha 50% workers get their jobs through networking )

HA: Alternative Hypothesis: p > 0.5 (More tha 50% workers get their jobs through networking ) (Claim)

Step 2:

n = 40

$\hat{p}$ = 27/40 = 0.675

n$\hat{p}$ = 40X 0.675 = 27 > 10

n (1 - $\hat{p}$ ) = 40 X 0.325 = 13 > 10

Both conditions are satisfied.

Step 3:

(i) Mean of sampling distribution = 0.675

(ii) Standard Deviation of sampling distribution is given by:

$\sqrt{\frac{0.5\times (1-0.5)}{40}}=0.0791$

Step 4:

(i) Z Test for one sample proportion

(ii)

$\alpha$ = 0.03

From Table, critical values of Z = 1.88

(iii)

Rejection Region:
Reject H0 if Z is greater than 1.88

Step 5:

Test Statistic is given by:

Z=- 0.675 -0.50 0.50 x (1 – 0.50)/40 = 2.214

Step 6:

Since calculated value of Z = 2.214 is greater than critical value of Z = 1.88, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that more tha 50% workers get their jobs through networking.

Let me know in the comment section if anything is not clear. I will reply ASAP!
If you liked the answer, please give a thumbs-up. This will be quite encouraging for me.Thank-you!