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Aleksandr-060686 [28]
3 years ago
5

Suppose you have a normally distributed set of data pertaining to a standardized test. The mean score is 500 and the standard de

viation is 100. What is the z-score of 450 point score?
Mathematics
1 answer:
Zinaida [17]3 years ago
4 0
I don't know the answer but I am sure you could go on Khan Academy to find the answer and a tutorial fully explaining how to solve it in the future.

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HeLp!!
erica [24]

Answer:

C

Step-by-step explanation:

A constant correlation is essentially when the points on a scatter plot do not show any type of pattern or correlation; the points are literally scattered rather randomly, which would cause the graph to neither increase nor decrease.

Meanwhile, a positive correlation means that the points on the plot follow a line with a positive slope. In other words, it increases to the right.

Thus, the answer is C.

Hope this helps!

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Y=x^2-6x+7 from algebra nation
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Answer:

Step-by-step explanation:

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ikadub [295]

Answer:

b

Step-by-step explanation:

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3 years ago
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Dylan Rieder is a statistics student investigating whether athletes have better balance than non-athletes for a thesis project.
Kobotan [32]

Answer:

t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457

p_v =P(t_{75}  

Comparing the p value with a significance level for example \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.  

Step-by-step explanation:

Data given and notation

\bar X_{A}=3.7 represent the mean for athletes  

\bar X_{NA}=4.1 represent the mean for non athletes  

s_{A}=1.1 represent the sample standard deviation for athletes  

s_{NA}=1.3 represent the sample standard deviation for non athletes

n_{A}=32 sample size for the group 2  

n_{NA}=45 sample size for the group 2  

\alpha=0.01 Significance level provided  

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the population mean for athletes is lower than the population mean for non athletes, the system of  hypothesis would be:  

Null hypothesis:\mu_{A}-\mu_{NA}\geq 0  

Alternative hypothesis:\mu_{A} - \mu_{NA}< 0  

We don't have the population standard deviation's, we can apply a t test to compare means, and the statistic is given by:  

t=\frac{(\bar X_{A}-\bar X_{NA})-\Delta}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{NA}}{n_{NA}}}} (1)  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

With the info given we can replace in formula (1) like this:  

t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457

P value  

We need to find first the degrees of freedom given by:

df=n_A +n_{NA}-2=32+45-2=75

Since is a one left tailed test the p value would be:  

p_v =P(t_{75}  

Comparing the p value with a significance level for example \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.  

5 0
3 years ago
A community sports league is raising money by making custom shirts to sell at league games. They plan to sell the shirts for $14
11111nata11111 [884]

Answer:

A. (14 – 7)n – 55

Step-by-step explanation:

it cost $14 to make so $14 - $7 to make then you have to advertise it AFTER and it takes more money to advertise so, -55 so the answer would be:

(14 – 7)n – 55

5 0
3 years ago
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