ASAP WILL GIVE THANKS!!!!!!!!! Solve the equation for h. h/20=5/4

The thicknesses of 81 randomly selected aluminum sheets were found to have a variance of 3.23. Construct the 98% confidence inte

Yuliya22 [10]

Answer:

The confidence interval for the population variance of the thicknesses of all aluminum sheets in this factory is Lower limit = 2.30, Upper limit = 4.83.

Step-by-step explanation:

The confidence interval for population variance is given as below:

$[(n - 1)\times S^{2} / X^{2} \alpha/2, n-1 ] < \alpha < [(n- 1)\times S^{2} / X^{2} 1- \alpha/2, n- 1 ]$

We are given

Confidence level = 98%

Sample size = n = 81

Degrees of freedom = n – 1 = 80

Sample Variance = S^2 = 3.23

$X^{2}_{[\alpha/2, n - 1]} = 112.3288\\\X^{2} _{1 -\alpha/2,n- 1} = 53.5401$

(By using chi-square table)

[(n – 1)*S^2 / X^2 α/2, n– 1 ] < σ^2 < [(n – 1)*S^2 / X^2 1 -α/2, n– 1 ]

[(81 – 1)* 3.23 / 112.3288] < σ^2 < [(81 – 1)* 3.23/ 53.5401]

2.3004 < σ^2 < 4.8263

Lower limit = 2.30

Upper limit = 4.83.

8 0

2 years ago

Solve simultaneously 2 log y=log 2 + log x 2 to power y = 4 to power x

skad [1K]

Answer:

Step-by-step explanation:

hello : here is an solution

3 0

3 years ago

Suppose the standard deviation of a normal population is known to be 3, and H0 asserts that the mean is equal to 12. A random sa

blagie [28]

Answer:

$z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9$

$p_v =P(z>1.9)=0.0287$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=12.95$ represent the sample mean

$\sigma=3$ represent the population standard deviation for the sample

$n=36$ sample size

$\mu_o =12$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 12, the system of hypothesis would be:

Null hypothesis: $\mu \leq 12$

Alternative hypothesis: $\mu > 12$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9$

P-value

Since is a right tailed test the p value would be:

$p_v =P(z>1.9)=0.0287$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.

4 0

3 years ago

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