Can someone help me please

Mathematics

2 answers:

miss Akunina [59]3 years ago

3 0

15 is A, 22+X=8
I don't have time to solve the rest RN but will come back later

hjlf3 years ago

3 0

Answer:

15. is D 16. is C 17. is A 18. is D

Step-by-step explanation:

Im in algebra 2 i know

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Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in

Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - α)% confidence interval for the population variance is given as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

It is provided that:

n = 20

s = 3.9

Confidence level = 95%

⇒ α = 0.05

Compute the critical values of Chi-square:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907$

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

$\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45$