Write the equation in standard form -5y 6y = -12

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.

8 0

2 years ago

Please help i will give brainliest!!

valina [46]

Answer:

1225miles

$\frac{(1225)}{(1.578) \times {(10}^{ - 5} )}$

8 0

3 years ago

A college entrance exam company determined that a score of 24 on the mathematics portion of the exam suggests that a student is

seraphim [82]

Answer:

a) The Z -value 2.397 < 2.576 at 99% or 0.01% level of significance

Null hypothesis is accepted at 0.01% level of significance

They score is above 24 on the math portion of the exam

b)

Null Hypothesis: There is no significance difference between the college level mathematics and math courses in high school

H₀: μ = 24

Alternative Hypothesis: H₁: μ ≠ 24

Step-by-step explanation:

Step(i):-

Given random sample 'n' = 250

Given data sample mean x⁻ = 24.5

Standard deviation = 3.3

Null Hypothesis: There is no significance difference between the college level mathematics and math courses in high school

H₀: μ = 24

Alternative Hypothesis: H₁: μ ≠ 24

test statistic

$Z = \frac{x^{-} - mean}{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{24.5 - 24}{\frac{3.3}{\sqrt{250} } } = \frac{0.5}{0.2087} = 2.397$

a) 99% or 0.01% level of significance

Level of significance ∝ = 0.01

$Z_{\frac{\alpha }{2} } = Z_{\frac{0.01}{2} } = Z_{0.005} =2.576$

The Z -value 2.397 < 2.576 at 99% or 0.01% level of significance

Null hypothesis is accepted at 0.01% level of significance

They score is above 24 on the math portion of the exam

b) 95% or 0.05% level of significance

Level of significance ∝ = 0.05

$Z_{\frac{\alpha }{2} } = Z_{\frac{0.05}{2} } = Z_{0.025} = 1.96$

The Z -value 2.397 > 1.96 at 95% or 0.05% level of significance

Null hypothesis is Rejected at 0.05% level of significance

They score is below 24 on the math portion of the exam

6 0

3 years ago