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stellarik [79]
3 years ago
6

Carla earned b bonus points. Shannon earned 24 more points than carla. Choose the expression that shows how many bonus points Sh

annon earned
Mathematics
1 answer:
True [87]3 years ago
8 0

Answer:6777

Step-by-step explanation: 1+6776=6777

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they moved forward 15 yards

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The correlation coefficient r between an employee’s age, x, and yearly salary, y, is 0.633. What percent of the variation in yea
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Find the measure of the angle indicated. <br> ?<br> 82<br> A) 77°<br> C) 82<br> B) 75<br> D) 127°
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c) 82

Step-by-step explanation:

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3 years ago
What is the solution to<br> 4(2x– 3)= 2(3x+ 1)?<br><br><br> -5<br><br> 1<br><br> 7<br><br> 10
Paul [167]
4 (2x – 3)= 2( 3x + 1)
<span>(4 (2x – 3)= 2( 3x + 1)) </span>÷ 2
2 (2x – 3) = (3x + 1)
4x - 6 = 3x +1
-3x       -3x
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3 years ago
In a​ survey, 32​% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result
adoni [48]

Answer:

z=\frac{0.29 -0.32}{\sqrt{\frac{0.32(1-0.32)}{210}}}=-0.932  

p_v =P(z  

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of  pet owners contacted by telephone is NOT significantly lower than 0.32

Step-by-step explanation:

Data given and notation

n=210 represent the random sample taken

X=61 represent the number of pet owners contacted by telephone

\hat p=\frac{61}{210}=0.29 estimated proportion of  pet owners contacted by telephone

p_o=0.32 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportions is lower than 0.32.:  

Null hypothesis:p \geq 0.32  

Alternative hypothesis:p < 0.32  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.29 -0.32}{\sqrt{\frac{0.32(1-0.32)}{210}}}=-0.932  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

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

p_v =P(z  

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of  pet owners contacted by telephone is NOT significantly lower than 0.32

3 0
3 years ago
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