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faltersainse [42]
3 years ago
13

A psychology professor wants to know whether verbal ability is related to memory quality in current first-year students at her s

mall college. Participants in the study (first-year students at her college) complete an online memory task. The students are first shown a list of 60 words. Next they are shown a list of 10 words that were on the original list. Then they are asked to identify the words on the second list that appeared on the original list. She uses the percentage of words that were correctly recognized on the original list as her measure of memory quality. She also asks the students to report several characteristics such as their age, gender, and verbal SAT score. Each of the 750 first-year students (338 males and 412 females) at her school volunteers to participate. The professor chose 75 students at random to complete the memory task and answer the questions. The average percentage of words that were correctly recognized on the original list was 68%. The professor infers that if all 750 first-year students had completed the study, the results would show that an average of 68% (plus or minus sampling error) of the words were correctly recognized as being on the original list. Which of the following are variables in the study? Check all that apply.
A. The students' verbal SAT scores
B. The students' percentage of words that were correctly recognized on the original list
C. The 75 students
D. The 750 students
Mathematics
1 answer:
alexgriva [62]3 years ago
8 0

Answer:

The  variable are

A and  B

Step-by-step explanation:

Generally we can define a variable as a name of a placeholder representing a value  now considering the option to be selected from we see that

    The students' verbal SAT scores is a variable  because it is a phrase or a place holder that represent a value (in the is case a  numerical value)  which the score its self

    The second option  The students' percentage of words that were correctly recognized on the original list is also a variable because it is a placeholder of a name (in this case a phrase ) that represented the actual value itself

  The  third option The 75 students is not a variable because it is not representing any value but itself is the value

  The  third option The 750 students is not a variable because it is not representing any value but itself is the value

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Electricity usage data consists of 45 months has a mean number of units consumed is 390.47 per month with a standard deviation o
neonofarm [45]

Answer:

The 95% confidence interval for the average monthly electricity consumed units is between 47.07 and 733.87

Step-by-step explanation:

We have the standard deviation for the sample. So we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 45 - 1 = 44

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 44 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 2.0141

The margin of error is:

M = T*s = 2.0141*170.5 = 343.4

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 390.47 - 343.40 = 47.07 units per month

The upper end of the interval is the sample mean added to M. So it is 390.47 + 343.40 = 733.87 units per month

The 95% confidence interval for the average monthly electricity consumed units is between 47.07 and 733.87

6 0
3 years ago
Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim
tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{10959}{25} =438.36

Sum of squares of differences = 175413.76

S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49

Population mean, μ = 425 mg/L

Sample mean, \bar{x} = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}

We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813

Now, t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108

Since,                        

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0
3 years ago
Researchers for a company that manufactures batteries want to test the hypothesis that the mean battery life of their new batter
guapka [62]

(b) one-sample t-test for a population mean

ur welcome :D

8 0
3 years ago
The least value of the function x^2 + px + q is 3 and this occurs when x = - 2. find the values of p and q
Rasek [7]

Answer:

Hello,

p=4

q=7

Step-by-step explanation:

The vertex of the parabola is (-2,3)

Equation of the parabola is  k(x+2)²+3=x²+px+q

Let's identify the coefficients:

kx²+4kx+4k+3=x²+px+q

so

k=1

4*1=p

4*1+3=q

3 0
2 years ago
Simplify the expression. Show your work.<br><br><img src="https://tex.z-dn.net/?f=-3%284a-5b%29" id="TexFormula1" title="-3(4a-5
neonofarm [45]

Step-by-step explanation:

-3 (4a-5b)

=-3×4a -(-3×5b)

=-12a+15b

5 0
4 years ago
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