What is the answer to this problem -3q+4=13

A researcher wants to find out if there is a significant difference between men and women in the average number of tattoos that

kifflom [539]

Answer:

The correct option is (A).

Step-by-step explanation:

The p-value is well-defined as per the probability, [under the null-hypothesis (H₀)], of attaining a result equivalent to or more extreme than what was the truly observed value of the test statistic.

A small p-value (typically p < 0.05) specifies strong evidence against the null hypothesis (H₀), so the null hypothesis is rejected. A p-value less than 0.05 is considered as statistically significant.

A large p-value (p > 0.05) specifies fragile proof against the H₀, so the null hypothesis is failed to be rejected.

In this case the p-value of the test is,

p-value = 0.003

And the significance level of the test was,

α = 0.05.

p-value = 0.003 < α = 0.05.

The p-value less than the significance level of the test.

Thus, the results are statistically significant.

The correct option is (A).

8 0

3 years ago

Use grouping symbols to make the following equation true. 5³ ÷ 5+20=5

Alona [7]

In this question,We have to group the given expression to get 5 as the final answer .

$5^3 \div 5+20$

Here we put 5 +20 in parenthesis, that is

$5^3 \div (5+20)$

And we do the required operations, that is

$=125 \div 25 =5$

And we get 5.

3 0

3 years ago

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:

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

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