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gladu [14]
3 years ago
6

A superintendent of schools wonders if scores on a test of science achievement differ for female and male 8th grade students. To

test this, she gathers a random sample of male 8th graders and a random sample of female 8th graders and compares their sample means. The test statistic that the superintendent should use to test her hypothesis is the ________.
a. 2-dependent sample t-test
b. 2-independent sample t-test
c. Correlation
d. 2-way ANOVA
Mathematics
1 answer:
lina2011 [118]3 years ago
5 0

Answer:

b. 2-independent sample t-test

Correct for this case is the best test since the result from each group are independent and we can compare the sample means between th two groups

Step-by-step explanation:

For this case the intention is try to test if scores on a test of science achievement differ for female and male 8th grade students. let's analyze the possible options for this case:

a. 2-dependent sample t-test

False is not possible since the nature of the gender is not possible to conclude tha the results from boys and girls are dependent

b. 2-independent sample t-test

Correct for this case is the best test since the result from each group are independent and we can compare the sample means between the two groups

c. Correlation

False we can't compare the means of interest with a correlation coefficient since that's not the purpose of the study

d. 2-way ANOVA

False we have just one variable between two groups so is not possible to apply a 2 way ANOVA

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The value below lies between which two integers on a number line.
I am Lyosha [343]

Answer:

2(29D-28adn-29)

Step-by-step explanation:

7 0
3 years ago
ACD is a triangle and B is a point on AC. AB = 8cm and BC is 6cm. Angle BCD = 48° and angle BDC = 50°. (a) Find the length of BD
FromTheMoon [43]

Answer:

  • 5.8206 cm
  • 10.528 cm
  • 23.056 cm^2

Step-by-step explanation:

(a) The Law of Sines can be used to find BD.

  BD/sin(48°) = BD/sin(50°)

  BD = (6 cm)(sin(48°)/sin(60°)) ≈ 5.82064 cm

__

(b) We can use the Law of Cosines to find AD.

  AD^2 = AB^2 +BD^2 -2·AB·BD·cos(98°) . . . . . angle ABD = 48°+50°

  AD^2 ≈ 110.841

  AD ≈ √110.841 ≈ 10.5281 . . . cm

__

(c) The area of ∆ABD can be found using the formula ...

  A = ab·sin(θ)/2 . . . . . where a=AB, b=BD, θ = 98°

  A = (8 cm)(5.82064 cm)sin(98°)/2 ≈ 23.0560 cm^2

_____

Angle ABD is the external angle of ∆BCD that is the sum of the remote interior angles BCD and BDC. Hence ∠ABD = 48° +50° = 98°.

5 0
3 years ago
4/10 + 40/100 = ? Help please
Kobotan [32]

Answer:

4/10 is the same as 40/100

there fore 4/10+40/100=40/100+40/100=

80/100

hope it helps :)

4 0
2 years ago
Read 2 more answers
Could anybody help me with this please?
KengaRu [80]
I think the opposite would be a circle because it would have any points.
4 0
3 years ago
Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7

s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262

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

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

7 0
3 years ago
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