Question

37. How do professionals stay on top of their​ careers? Of 1,000 surveyed members of a business networking​ website, 570 reported that they engaged in professional networking within the last month. At the 0.05 level of​ significance, is there evidence that the proportion of all site members who engaged in professional networking within the last month is different from 54​%?

State the null and alternative hypotheses.

Calculate the test statistic.

What is the​ p-value?

38. A consulting group recently conducted a global survey of product teams with the goal of better understanding the dynamics of product team performance and uncovering the practices that make these teams successful. One of the survey findings was that 34​% of organizations have a coherent business strategy that they stick to and effectively communicate. Suppose another study is conducted to check the validity of this​ result, with the goal of proving that the percentage is less than 34​%.

a. State the null and research hypotheses.

b. A sample of 200 organizations is​ selected, and results indicate that 61 organizations have a coherent business strategy that they stick to and effectively communicate. Use the​ p-value approach to determine at the 0.05 level of significance whether there is evidence that the percentage is less than 34​%.

Calculate the test statistic.

Calculate the​ p-value.

Answer 1

37)

P= 0.54, n= 1000, x =570, =0.05

hence null and alternative hypothesis is as follows,

Ho: P = 0.54

H1: P 0.54 (Claim)

Calculate tests statistics

z= 1.903

test statistics = 1.903

now calculate p-value for two tailed test

P-Value = 2 * 1 - P(z < 1.903)

find P(z < 1.903) using z table we get

P(z < 1.903) = 0.9713

P-Value = 2 * 1 - P(z < 1.903)

P-Value = 2 * 1 - 0.9713

P-Value = 0.0574

since ( p-value =0.0574) > ( =0.05)

Failed to reject null hypothesis

Therefore there is not sufficient evidence to support the claim that the proportion of all site members who engaged in professional networking within the last month is different from 54​%.

38)

a)

null and research hypothesis is as follows,

Ho: P = 0.34

H1: P <  0.34

b)

n=200, x=61, =0.05

Calculate tests statistics

z= −1.045

test statistics = −1.045

now calculate p-value for two tailed test

P-Value = P(z < −1.045)

find P(z < −1.045) using z table we get

P(z < −1.045) = 0.1480

P-Value = 0.1480

since ( p-value =0.1480) > ( =0.05)

Failed to reject null hypothesis

Therefore there is not sufficient evidence to support the claim hat the percentage is less than 34​%.