The recruiting firm Jobvite, Inc. reported that 78% of U.S. companies use social
ID: 3227627 • Letter: T
Question
The recruiting firm Jobvite, Inc. reported that 78% of U.S. companies use social networks such as Facebook and LinkedIn to recruit job candidates. An economist thinks that the percentage is higher at technology companies. She samples 70 technology companies in a simple random sample and finds that 55 of them use social networks. Can she conclude that more than 78% of technology companies use social networks to recruit job candidates? Claim in symbolic form: ______ Opposite in symbolic form: ______ H_0 in symbolic: ________ H_1 in symbolic form: _______ Which curve is appropriate? (circle one) Degrees of freedom (if not applicable, write NONE on the blank) df = ______ What is the tail structure of the test? (circle one) left right two What is (are) the critical value(s)? _____ What is the value of the test statistic? _______ What is the P-value? _______Explanation / Answer
SOlution:-
1) Claim in symbolic form: More than 78% technology companies use social network to recruit job candidates.
2) Opposite in symbolic form: Less than or equal to 78% technology companies use social network to recruit job candidates.
3) H 0: Null hypothesis: P < 0.78
4) H 1: Alternative hypothesis: P > 0.78(Our claim)
5) Z curve would be appropriate, since n is quite large.
6) n p > 5
7) n(1-p) > 5
8) Degree of freedom is not applicable.
9) The test is Right tailed test.
10) The critical value is 1.96
11) The value of the test statistic is 0.1154
12) The p value is 0.452
13) Fail to reject the H0
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.78
Alternative hypothesis: P > 0.78
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.04951
z = (p - P) /
z = 0.1154
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the z-score is more than 0.1154. We use the Normal Distribution Calculator to find P(z > 0.1154) = 0.452
Interpret results. Since the P-value (0.452) is greater than the significance level (0.05), we have to accept the null hypothesis.
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