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2 years ago

Pllllllllzzzzzzz answer this!!!♥️♥️

Mathematics

1 answer:

Umnica [9.8K]2 years ago

6 0

Answer:

Use the regression calculator to compare the teams’ number of runs with their number of wins.

A 2-column table with 9 rows. Column 1 is labeled R with entries 808, 768, 655, 684, 637, 619, 613, 609, 563. Column 2 is labeled W with entries 93, 94, 66, 81, 86, 75, 61, 69, 55.

What is the y-intercept of the trend line, to the nearest hundredth?

Step-by-step explanation:

Guest

1 year ago

Dude , you are suppose to answer it not question it!

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Calculate the monthly payment for a 5-year car loan of $23,570 at 10.43% interest, compounded monthly. A. $247.44 b. $337.56 c.

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Jay earns $5.90 per hour working at the ice cream parlor after school. He needs at least $236 for a stereo system. Choose the in

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This is because Jay needs at least 236 therefore the money he earns from working at the ice cream parlor would need to be higher than 236.

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The confidence interval for the difference in proportions is

$-0.028\leq p_1-p_2 \leq 0.096$

No. As the 95% CI include both negative and positive values, no proportion is significantly different from the other to conclude there is a difference between them.

Step-by-step explanation:

We have to construct a confidence interval for the difference of proportions.

The difference in the sample proportions is:

$p_1-p_2=x_1/n_1-x_2/n_2=(183/217)-(322/398)=0.843-0.809\\\\p_1-p_2=0.034$

The estimated standard error is:

$\sigma_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} } \\\\\sigma_{p_1-p_2}=\sqrt{\frac{0.843*0.157}{217}+\frac{0.809*0.191}{398} } \\\\\sigma_{p_1-p_2}=\sqrt{0.000609912+0.000388239}=\sqrt{0.000998151} \\\\ \sigma_{p_1-p_2}=0.0316$

The z-value for a 95% confidence interval is z=1.96.

Then, the lower and upper bounds are:

$LL=(p_1-p_2)-z*\sigma_p=0.034-1.96*0.0316=0.034-0.062=-0.028\\\\\\UL=(p_1-p_2)+z*\sigma_p=0.034+1.96*0.0316=0.034+0.062=0.096$

The confidence interval for the difference in proportions is

$-0.028\leq p_1-p_2 \leq 0.096$

<em>Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?</em>

No. It can not be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group, as the confidence interval include both positive and negative values.

This means that we are not confident that the actual difference of proportions is positive or negative. No proportion is significantly different from the other to conclude there is a difference.

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