Question

1.A recent article reported that a job awaits only one in three new college graduates. (1 in 3 means the proportion is .333) A survey of 200 recent graduates revealed that 80 graduates had jobs. At the .02 significance level, we will conduct a hypothesis test to determine if we can conclude if a larger proportion of graduates have jobs than previously reported. What will be the value of our critical value?

2..A recent article reported that a job awaits only one in three new college graduates. A survey of 200 recent graduates revealed that 80 graduates had jobs. At the .02 significance level, we will conduct a hypothesis test to determine if we can conclude if a larger proportion of graduates have jobs than reported. What will be the value of our test statistic?

3.. A recent article reported that a job awaits only one in three new college graduates. A survey of 200 recent graduates revealed that 80 graduates had jobs. At the .02 significance level, can we conclude that a larger proportion of graduates have jobs than reported in the article? Yes/ No

4.. Given the hypothesis Ho: u=400, you conduct a sample of 12 observations and obtain a sample mean of 407 and a sample standard deviation of 6. Using a .01 significance level to test to see if you would reject Ho, what would be the value of your critical value? i.e the positive one?

Answer 1

1) NULL HYPOTHESIS H0: $P=0.333$

ALTERNATIVE HYPOTHESIS HA:

alpha= 0.02

p= x/n= 80/200= 0.4

$Z= p-P/\sqrt(P*Q/n)$

Z 0.4-0.333/ V/(0.333 * 0.667/200)

$Z= 0.067/\sqrt(0.333*0.667/200)$

$Z= 0.067/\sqrt(0.222111/200)$

Test Statistic

Z critical value FOR ONE TAILED TEST= 2.05

CONCLUSION: SINCE CRITICAL VALUE OF Z IS GREATER THAN CALCULATED VALUE OF Z . WE THEREFORE DO NOT REJECT NULL HYPOTHESIS AT 0.02 LEVEL OF SIGNIFICANCE.

3) We can not conclude that a larger proportion of graduates have jobs than reported in the article.

4) $H0: \mu=400$

$Ha: \mu\ne400$

alpha= 0.01

n=12 , xbar= 407 and standard deviation= 6

Under null hypothesis test statistic is

$t= \overline{x}-\mu/s/\sqrt(n)$

$t= 407-400/6/\sqrt(12)$

degrees of freedom= n-1= 12-1= 11

Critical value t= 3.11 for two tailed test. Yes positive one.

Conclusion: Since critical value of t is smaller than calculated value of t we therefore reject null hypothesis at 0.01 level of significance.