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marin [14]
4 years ago
8

An EASY QUESTION because I’m dumb sometimes 30 POINTS!( 05.05 LC)

Mathematics
1 answer:
vaieri [72.5K]4 years ago
3 0

Answer:

Wonder Woman

Step-by-step explanation:

she has the highest total

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When I sit and watch my students take exams, I often think to myself "I wonder if students with bright calculators are impacted
hodyreva [135]

Answer:

There is a difference between the two means.

Step-by-step explanation:

The hypothesis can be defined as:

<em>H₀</em>: The mean exam scores of my SAT 215 students with colorful calculators are same as the mean scores of my STA 215 students with plain black calculators, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.

<em>Hₐ</em>: The mean exam scores of my SAT 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 0.

Assume that the significance level of the test is, <em>α</em> = 0.05. Also assuming that the population variances are equal.

The decision rule:

A 95% confidence interval for mean difference can be used to determine the result of the hypothesis test. If the 95% confidence interval contains the null hypothesis value, i.e. 0 then the null hypothesis will not be rejected.

The 95% confidence interval for mean difference is:

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

Compute the pooled standard deviation as follows:

S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}} {n_{1}+n_{2}-2}}}=\sqrt{\frac{(49-1)(4.7)^{2}+(38-1)(5.7)^{2}}{49+38-2}}=5.16

The critical value of <em>t</em> is:

t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (49+38-2)}=t_{0.025, 85}=1.984

*Use a <em>t</em>-table.

Compute the 95% confidence interval for mean difference as follows:

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

     =(84-87)\pm 1.984\times 5.16\times \sqrt{\frac{1}{49}+\frac{1}{38}}

     =-3\pm 2.133\\=(-5.133, -0.867)\\\approx(-5.13, -0.87)

The 95% confidence interval for mean difference is (-5.13, -0.87).

The confidence interval does not contains the value 0. This implies that the null hypothesis will be rejected at 5% level of significance.

Hence, concluding that the mean exam scores of my STA 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators.

7 0
4 years ago
Not enough mashed potatoes for thanksgiving?
LenaWriter [7]

what? idk, this is int even math.

6 0
3 years ago
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Help please!!<br><br>x/2=x+6/4​
Goshia [24]

Answer:

-4 I think

Step-by-step explanation:

7 0
3 years ago
Seams Personal advertises on its website that 95% of customer orders are received within four working days. They performed an au
Anna71 [15]

Answer:

500 orders

Step-by-step explanation:

500

4 0
3 years ago
1 slope y-intercept equation o 6 5 4 3 2 2 3 5 j-intercept​
mars1129 [50]

Answer:

Step-by-step explanation:

(-1,1) and (0,-2) are points on the line

slope of line = (-2-1)/(0-(-1)) = -3

y-intercept = -2

y = -3x - 2

3 0
3 years ago
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