If the graph of a line goes from the upper left corner of the graph to the lower right corner, what can you say about its slope?

1 answer:

4 0

You can say that the slope is negative. A line on a graph that goes from the upper left corner to the bottom right is a negative slope.

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One of your peers claims that boys do better in math classes than girls. Together you run two independent simple random samples

JulijaS [17]

Answer:

Step-by-step explanation:

Hello!

To test if boys are better in math classes than girls two random samples were taken:

Sample 1

X₁: score of a boy in calculus

n₁= 15

X[bar]₁= 82.3%

S₁= 5.6%

Sample 2

X₂: Score in the calculus of a girl

n₂= 12

X[bar]₂= 81.2%

S₂= 6.7%

To estimate per CI the difference between the mean percentage that boys obtained in calculus and the mean percentage that girls obtained in calculus, you need that both variables of interest come from normal populations.

To be able to use a pooled variance t-test you have to also assume that the population variances, although unknown, are equal.

Then you can calculate the interval as:

[(X[bar]_1-X[bar_2) ± $t_{n_1+n_2-2;1-\alpha /2}$ * $Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$ ]

$Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} } = \sqrt{\frac{14*(5.6^2)+11*(6.7^2)}{15+12-2} }= 6.108= 6.11$

$t_{n_1+n_2-2;1-\alpha /2}= t_{15+12-2;1-0.05}= t_{25;0.95}= 1.708$

[(82.3-81.2) ± 1.708* (6.11* $\sqrt{\frac{1}{15}+\frac{1}{12} }$ ]

[-2.94; 5.14]

Using a 90% confidence level you'd expect the interval [-2.94; 5.14] to contain the true value of the difference between the average percentage obtained in calculus by boys and the average percentage obtained in calculus by girls.

I hope this helps!

3 0

3 years ago

This is super hard help and I will give a thnxs

Sladkaya [172]

42|2
21|3
7|7
1

The combination is 237.

3 0

3 years ago

Read 2 more answers

Hey anybody plz help me ;C

jekas [21]

Answer: