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Rasek [7]
2 years ago
7

1 of 6 The length of each side of a square is 6.1 cm rounded to 1 decimal place. Write the error interval for the perimeter, p ,

in the form a ≤ p < b .
Mathematics
1 answer:
arsen [322]2 years ago
8 0

The error interval of the perimeter is 6.1 cm ≤ p < 6.15 cm

<h3>How to determine the error of interval?</h3>

The perimeter is given as:

P = 6.1 cm to 1 decimal place

The smallest number that can be approximated to 6.1 is 6.05, while the largest number is less than 6.15

This means that the interval is:

6.1 cm ≤ p < 6.15

Read more about error interval at:

brainly.com/question/16810055

#SPJ1

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z=\frac{0.884 -0.72}{\sqrt{\frac{0.72(1-0.72)}{43}}}=2.395  

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Since the p value is lower than the significance level of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of Americans that they would prefer an American automobile  is significantly different from 0.72 and then the claim is correct

Step-by-step explanation:

Information given

n=43 represent the random sample taken

X=38 represent the Americans who they would prefer an American automobile

\hat p=\frac{38}{43}=0.884 estimated proportion of Americans that they would prefer an American automobile

p_o=0.72 is the value that we want to check

\alpha=0.1 represent the significance level

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to verify if the percentage of Americans that they would prefer an American automobile  today differs from 72%, then the system of hypothesis are:

Null hypothesis:p=0.72  

Alternative hypothesis:p \neq 0.72  

The statistic is given by:

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

Replacing the data given we got:

z=\frac{0.884 -0.72}{\sqrt{\frac{0.72(1-0.72)}{43}}}=2.395  

Now we can find the p value taking incount that we are using a bilateral test

p_v =2*P(z>2.395)=0.01662  

Since the p value is lower than the significance level of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of Americans that they would prefer an American automobile  is significantly different from 0.72 and then the claim is correct

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3 years ago
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