I need help, please and thank you.

On Andy's swim team there are 8 girls and 12 boys. What percent of the team are boys?

vitfil [10]

Answer:

63% maybe I'm not positive on that

3 0

3 years ago

35% of what number is 28

Finger [1]

The answer to your question is 80

5 0

4 years ago

Use the t-distribution to find a confidence interval for a difference in means μ1-μ2 given the relevant sample results. Give the

Serggg [28]

Answer:

(a) The best estimate of $\mu_{1}-\mu_{2}$ is 13.2.

(b) The margin of error is 5.30.

(c) The 95% confidence interval for the difference between two means is (7.90, 18.50).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means using a t-interval is:

$CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n_{1}+n_{2}-2)} \sqrt{S_{p}^{2}[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}$

(a)

A point estimate of a parameter (population) is a distinct value used for the estimation the parameter (population). For instance, the sample mean $\bar x$ is a point-estimate of the population mean μ.

Similarly the point estimate of the difference between two means is:

$\bar x_{1}-\bar x_{2}$

Compute the point estimate of $\mu_{1}-\mu_{2}$ as follows:

$E(\mu_{1}-\mu_{2})=\bar x_{1}-\bar x_{2}\\=79.0-65.8\\=13.2$

Thus, the best estimate of $\mu_{1}-\mu_{2}$ is 13.2.

(b)

Compute the pooled variance as follows:

$S_{p}^{2}=\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}=\frac{(35-1)10.5^{2}+(20-1)7.2^{2}}{35+20-2}=89.311$

Compute the critical value of t as follows:

$t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (35+20-2)}=t_{0.025, 53}=2.00$

*Use a t-table.

Compute the margin of error as follows:

$MOE=t_{\alpha/2, (n_{1}+n_{2}-2)} \sqrt{S_{p}^{2}[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}$

$=2.00\times\sqrt{89.311\times [\frac{1}{35}+\frac{1}{20}]} \\=5.298\\\approx5.30$

Thus, the margin of error is 5.30.

(c)

Compute the 95% confidence interval for the difference between two means as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm MOE$

$=13.2\pm 5.298\\=(7.902, 18.498)\\\approx (7.90, 18.50)$

Thus, the 95% confidence interval for the difference between two means is (7.90, 18.50).

4 0

3 years ago

Solve the number sentences.write the double fact that helped you solve the double plus 1.

ASHA 777 [7]

1+1=2 2+1=3 (they are double plus 1 expressions

8 0

4 years ago

4 hours a week for a month. how many hours you have

love history [14]

7 days a week

4x7=28

About 4 weeks each month

28x4=112

The answer should be 112 hours. I hope this helped

7 0

4 years ago

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