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Anastaziya [24]
1 year ago
7

An engineer measures the peak current (in microamps) when a solution containing an amount of nickel (in parts per 106 ) is added

to a buffer. The experiment was repeated for eleven different values of nickel solutions. A scatterplot showing the data is given below: Suppose a regression line was added to the plot above. If an additional measurement had been taken with the nickel of 62 parts per 106 and a peak current of 0.38 microamps, adding this observation would: A. decrease the intercept, decrease the slope. B. decrease the intercept, increase the slope. C. increase the intercept, increase the slope. D. increase the intercept, decrease the slope.
Mathematics
1 answer:
dangina [55]1 year ago
7 0

Answer:

B. decrease the intercept, increase the slope

Step-by-step explanation:

A slope indicates the steepness of a line while the intercept points the location where it intercepts its axis. The linear relationship between can be defined using the intercept and the slope. Both concepts are used to estimate the average range of change. Since we are trying to add a peak current value of 0.38 which is lesser than the average, the intercept of the graph would therefore decrease and the slope increase.

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The number of newly reported crime cases in a county in New York State is shown in the accompanying table, where x represents th
Novay_Z [31]

Answer:

Step-by-step explanation:

I use  84+ CE

stat edit, then fill in the #s

then

vars 5

then

2'nd stat plot, on

then, click stat

Click arrow 1 time to the left to get to Calc

then click (4)(LinReg(ax+b))

then click enter 5 times

(y=-25.31428571x+1000.285714

y=-25.3x+1000.3

now, lets use computer:

y=-25.31(543)+1000.3

y=-12743.03

round to the biggest whole number )

this doesn't really work, so I will put 1999, 2000, 2001, 2002, 2003, 2004 instead of 0, 1, 2, 3, 4, 5 and do the same thing

now I get

y=-25.31428571x+51603.54286

y=-25.3x+51603.5

now, lets use computer:

y=-25.3(543)+51603.5

y=37865.6

round to the biggest whole number:

y=37866

so, year 37866

7 0
2 years ago
A shoe manufacturer compared material A and material B for the soles of shoes. Twelve volunteers each got two shoes. The left wa
sveta [45]

Answer:

a) Are dependent since we are mesuring at the same individuals but on different times and with a different method

b) If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

c) p_v =P(t_{(12)}>0.969) =0.353

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

The Q-Q plot, or quantile-quantile plot, "is a graphical tool to help us assess if a set of data plausibly came from some theoretical distribution such as a Normal or exponential".

Let put some notation  

x=value for A , y = value for B

A: 379, 378, 328, 372, 325, 304, 356, 309, 354, 318, 355, 392

B: 372, 376, 328, 368, 283, 252, 369, 321, 379, 303, 328, 411

(a) Are the two samples paired or independent? Explain your answer.

Are dependent since we are mesuring at the same individuals but on different times and with a different method

(b) Make a normal QQ plot of the differences within each pair. Is it reasonable to assume a normal population of differences?

The first step is calculate the difference d_i=A_i-B_i and we obtain this:

d: 7,2,0,4,42,52,-13,-12,-25,15,27,-19

In order to do the qqplot we can use the following R code:

d<-c(7,2,0,4,42,52,-13,-12,-25,15,27,-19)

qqnorm(d)

And the graph obtained is attached.

If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

(c) Choose a test appropriate for the hypotheses above and justify your choice based on your answers to parts (a) and (b). Perform the test by computing a p-value, make a test decision, and state your conclusion in the context of the problem

The system of hypothesis for this case are:

Null hypothesis: \mu_A- \mu_B = 0

Alternative hypothesis: \mu_A -\mu_B \neq 0

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{80}{12}=6.67

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =23.849

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{6.67 -0}{\frac{23.849}{\sqrt{12}}}=0.969

The next step is calculate the degrees of freedom given by:

df=n-1=12-1=11

Now we can calculate the p value, since we have a two tailed test the p value is given by:

p_v =P(t_{(12)}>0.969) =0.353

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

4 0
2 years ago
Kaitlyn gave 1/2 of her candy bar to arianna. arianna gave 1/3 of the candy she got from kaitlyn to cameron. what fraction of a
sukhopar [10]
Let the total candy bar be represented by x.

Amount of chocolate Kaitlyn gave to Arianna = \frac{1}{2}x

Amount of chocolate Arianna gave Cameron 

= \frac{1}{3} rd <span>of the candy she got from Kaitlyn to Cameron

= </span>\frac{1}{3} × \frac{1}{2}x

= \frac{1}{3}* \frac{1}{2} x  =  \frac{1}{6} x

Hence, Cameron got \frac{1}{6} th of the original chocolate.
5 0
2 years ago
Suppose that 90% of all dialysis patients will survive for at least 5 years. In a simple random sample of 100 new dialysis patie
Keith_Richards [23]

Answer:

The  probability   P(\^ p  >  0.80) =  0.99957

Step-by-step explanation:

From the question we are told that

     The population proportion is  p  =  0.90

     The sample size is  n  =  30

  Generally mean of the sampling distribution is  \mu_{\= x } =  p =  0.90

Generally the standard deviation is mathematically represented as

      \sigma  = \sqrt{\frac{p( -p )}{n} }

=>    \sigma  = \sqrt{\frac{0.90( 1 -0.90 )}{100} }

=>    \sigma  = 0.03

Generally the he probability that the proportion surviving for at least five years will exceed 80%, rounded to 5 decimal places is mathematically represented as

       P(\^ p  >  0.80) =  P(\frac{ \^ p - p }{ \sigma }  >  \frac{0.80 - 0.90}{0.03} )

Generally \frac{\^p - p}{\sigma}  =  Z(The standardized \  value  \  of  \^  p)

So  

    P(\^ p  >  0.80) =  P(Z  >  -3.33 )

From the z-table P(Z  >  -3.33 ) = 0.99957

So

    P(\^ p  >  0.80) =  0.99957

5 0
2 years ago
What is the final amount if 865 is decreased by 16% followed by a 14% increase?
Harman [31]

Answer:

828.32

Step-by-step explanation:

865 x (1 - 0.16) x (1 + 0.14) = 828.32

6 0
2 years ago
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