Answer:
Since the calculated value of z= 2.82 does not lie in the critical region the null hypothesis is accepted and it is concluded that the sample data support the authors' conclusion that the proportion of the country's boys who listen to music at high volume is greater than this proportion for the country's girls.
The value of p is 0 .00233. The result is significant at p < 0.10.
Step-by-step explanation:
1) Let the null and alternate hypothesis be
H0: μboys − μgirls > 0
against the claim
Ha: μboys − μgirls ≤ 0
2) The significance level is set at 0.01
3) The critical region is z ≤ ± 1.28
4) The test statistic
Z= p1-p2/ sqrt [pcqc( 1/n1+ 1/n2)]
Here p1= 397/768= 0.5169 and p2= 331/745=0.4429
pc = 397+331/768+745
pc= 0.4811
qc= 1-pc= 1-0.4811=0.5188
5) Calculations
Z= p1-p2/ sqrt [pcqc( 1/n1+ 1/n2)]
z= 0.5169-0.4429/√ 0.4811*0.5188( 1/768+ 1/745)
z= 2.82
6) Conclusion
Since the calculated value of z= 2.82 does not lie in the critical region the null hypothesis is accepted and it is concluded that the sample data support the authors' conclusion that the proportion of the country's boys who listen to music at high volume is greater than this proportion for the country's girls.
7)
The value of p is 0 .00233. The result is significant at p < 0.10.
Break into 2 equatins to get x's:
2x-6<4 and 2x-6>-4
Solve for x in both:
2x-6<4
2x<10
x<5
2x-6>-4
2x>2
x>1
Graph both inequalities on the same number line:
<---0-----1-----2-----3-----4-----5-----6--->
Answer:
It should be b
Step-by-step explanation:
I took this test
Answer:
The z-score for SAT exam of junior is much small than his ACT score. This means he performed well in his ACT exam and performed poor in his SAT exam.
Step-by-step explanation:
Mean SAT scores = 1026
Standard Deviation = 209
Mean ACT score = 20.8
Standard Deviation = 4.8
We are given SAT and ACT scores of a student and we have to compare them. We cannot compare them directly so we have to Normalize them i.e. convert them into such a form that we can compare the numbers in a meaningful manner. The best way out is to convert both the values into their equivalent z-scores and then do the comparison. Comparison of equivalent z-scores will tell us which score is higher and which is lower.
The formula to calculate the z-score is:
Here, μ is the mean and σ is the standard deviation. x is the value we want to convert to z score.
z-score for junior scoring 860 in SAT exam will be:
z-score for junior scoring 16 in ACT exam will be:
The z-score for SAT exam of junior is much small than his ACT score. This means he performed well in his ACT exam and performed poor in his SAT exam.