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Ket [755]
3 years ago
15

Researchers at Indiana University at Bloomington have found that chocolate milk effectively helps athletes recover from an inten

se workout. They had nine cyclists bike, rest four hours, then bike again, three separate times. After each workout, the cyclists downed chocolate milk or energy drinks Gatorade or Endurox (two to three glasses per hour); then, in the second workout of each set, they cycled to exhaustion. When they drank chocolate milk, the amount of time they could cycle until they were exhausted was similar to when they drank Gatorade and longer than when they drank Endurox.
Mathematics
1 answer:
victus00 [196]3 years ago
8 0

Answer:

The question is not complete. I have found online the following:

<em>Textbook Problem: </em>

<em>Researchers at Indiana University at Bloomington have found that chocolate milk effectively helps athletes recover from an intense workout. They had nine cyclists bike, rest four hours, then bike again, three separate times. After each workout, the cyclists downed chocolate milk or energy drinks Gatorade or Endurox (two to three glasses per hour); then, in the second workout of each set, they cycled to exhaustion. When they drank chocolate milk, the amount of time they could cycle until they were exhausted was similar to when they drank Gatorade and longer than when they drank Endurox. </em>

<em>To Determine: </em>

<em>Explain whether the researcher needs to use random assignment for the conclusion to be valid.</em>

Answer: Yes, random assignment is required.

Step-by-step explanation:

Random assignment is a technique in experiments to assign subject or elements to different groups randomly BEFORE beginning with the experiment processes. This allows to assume that each group's attributes are practically equivalent, and if the experiment produces different effects between the groups, it is not related to the specific individuals in the groups.

So to conclude that Chocolate Milk, Gatorade and Endurox have different effects, the cyclists groups must be assigned randomly.

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stiks02 [169]

Answer:

Q2. Original 100%

New Selling price=100% - 25% = 75% of original

Sale price of bracelet= 75/100 × 44 = $33

Q3. 100%- 25%= 75%

Cost of table= 75/100 × 425= $318.75

Alternatively,

Discount= 25/100 × 425 = $106.25

Cost of table= $425- $106.25= $318.75

Q4. 10 cans ---- $4

1 can= 4÷10 = $0.40

15 cans= $0.40 ×15 = $6

Q5. Total number of children= 35+5= 40

Total people= 10+40 = 50

Thus, the ratio is 40:50= 4:5 (<em>divide by 10</em>)

3 0
3 years ago
Choose the correct answer 1 1/12 - 7/10 greater than half or less than half​
CaHeK987 [17]

Answer:

less than half

Step-by-step explanation:

13/12 - 7/10 = 23/60

5 0
3 years ago
Is it true or false that the measure of A is 45°
GrogVix [38]
The answer is false
8 0
3 years ago
14.914 cm + 243.1 cm + 243.1 cm =269.0186 cm
Alexeev081 [22]

Answer:

False

Step-by-step explanation:

3 0
3 years ago
The principal at Crest Middle School, which enrolls only sixth-grade students and seventh-grade students, is interested in deter
AlekseyPX

Answer:

a) [ -27.208 , -12.192 ]

b) New procedure is not recommended

Step-by-step explanation:

Solution:-

- It is much more common for a statistical analyst to be interested in the difference between means than in the specific values of the means themselves.

- The principal at Crest Middle School collects data on how much time students at that school spend on homework each night.  

- He/She takes a " random " sample of n = 20 from a sixth and seventh grades students from the school population to conduct a statistical analysis.

- The summary of sample mean ( x1 & x2 ) and sample standard deviation ( s1 & s2 ) of the amount of time spent on homework each night (in minutes) for each grade of students is given below:

                                                          <u>Mean ( xi )</u>       <u> Standard deviation ( si )</u>

          Sixth grade students                 27.3                            10.8                  

          Seventh grade students           47.0                             12.4

- We will first check the normality of sample distributions.

  • We see that sample are "randomly" selected.
  • The mean times are independent for each group
  • The groups are selected independent " sixth " and " seventh" grades.
  • The means of both groups are conforms to 10% condition of normality.

Hence, we will assume that the samples are normally distributed.

- We are to construct a 95% confidence interval for the difference in means ( u1 and u2 ).

- Under the assumption of normality we have the following assumptions for difference in mean of independent populations:

  • Population mean of 6th grade ( u1 ) ≈ sample mean of 6th grade ( x1 )  
  • Population mean of 7th grade ( u2 ) ≈ sample mean of 6th grade ( x2 )

Therefore, the difference in population mean has the following mean ( u ^ ):

                      u^ = u1 - u2 = x1 - x2

                      u^ = 27.3 - 47.0

                      u^ = -19.7

- Similarly, we will estimate the standard deviation (Standard Error) for a population ( σ^ ) represented by difference in mean. The appropriate relation for point estimation of standard deviation of difference in means is given below:

                    σ^ =  √ [ ( σ1 ^2 / n1 ) + ( σ2 ^2 / n2 ) ]

Where,

          σ1 ^2 : The population variance for sixth grade student.

          σ2 ^2 : The population variance for sixth grade student.

          n1 = n2 = n : The sample size taken from both populations.

Therefore,

                 σ^ =  √ [ ( 2*σ1 ^2   / n )].

- Here we will assume equal population variances : σ1 ≈ σ2 ≈ σ is "unknown". We can reasonably assume the variation in students in general for the different grade remains somewhat constant owing to other reasons and the same pattern is observed across.

- The estimated standard deviation ( σ^ ) of difference in means is given by:

σ^ =

           s_p*\sqrt{\frac{1}{n_1} + \frac{1}{n_2}  } = s_p*\sqrt{\frac{1}{n} + \frac{1}{n}  } = s_p*\sqrt{\frac{2}{n}}\\\\\\s_p = \sqrt{\frac{(n_1 - 1 )*s_1^2 + (n_2 - 1 )*s_2^2}{n_1+n_2-2} } =  \sqrt{\frac{(n - 1 )*s_1^2 + (n - 1 )*s_2^2}{n+n-2} } = \sqrt{\frac{(n - 1 )*s_1^2 + (n - 1 )*s_2^2}{2n-2} } \\\\s_p = \sqrt{\frac{(20 - 1 )*s_1^2 + (20 - 1 )*s_2^2}{2(20)-2} } \\\\s_p = \sqrt{\frac{19*10.8^2 + 19*12.4^2}{38} } = \sqrt{135.2}  \\\\s_p = 11.62755

           σ^ = 11.62755*√2/20

          σ^ = 3.67695

- Now we will determine the critical value associated with Confidence interval ( CI ) which is defined by the standard probability of significance level ( α ). Such that:

         Significance Level ( α ) = 1 - CI = 1 - 0.95 = 0.05

                   

- The reasonable distribution ( T or Z ) would be determined on the basis of following conditions:

  • The population variances ( σ1 ≈ σ2 ≈ σ )  are unknown.
  • The sample sizes ( n1 & n2 ) are < 30.

Hence, the above two conditions specify the use of T distribution critical value. The degree of freedom ( v ) for the given statistics is given by:

          v = n1 + n2 - 2 = 2n - 2 = 2*20 - 2

          v = 38 degrees of freedom        

- The t-critical value is defined by the half of significance level ( α / 2 ) and degree of freedom ( v ) as follows:

          t-critical = t_α / 2, v = t_0.025,38 = 2.024

- Then construct the interval for 95% confidence as follows:

          [ u^ - t-critical*σ^ , u^ + t-critical*σ^ ]

          [ -19.7 - 2.042*3.67695 , -19.7 + 2.042*3.67695 ]

          [ -19.7 - 7.5083319 , -19.7 + 7.5083319 ]

          [ -27.208 , -12.192 ]

- The principal should be 95% confident that the difference in mean times spent of homework for ALL 6th and 7th grade students in this school (population) lies between: [ -27.208 , -12.192 ]

- The procedure that the matched-pairs confidence interval for the mean difference in time spent on homework prescribes the integration of time across different sample groups.

- If we integrate the times of students of different grades we would have to  make further assumptions like:

  • The intelligence levels of different grade students are same
  • The aptitude of students from different grades are the same
  • The efficiency of different grades are the same.

- We have to see that both samples are inherently different and must be treated as separate independent groups. Therefore, the above added assumptions are not justified to be used for the given statistics. The procedure would be more bias; hence, not recommended.

                 

8 0
3 years ago
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