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Marrrta [24]
3 years ago
6

Avanety of two types of snack packs are delivered to a store. The box plots compare the number of calories in each

Mathematics
2 answers:
AveGali [126]3 years ago
7 0

Answer:

The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs

of crackers

Step-by-step explanation:

IQR of trail mix data = 105 - 90 = 15

The range of cracker data = 100 - 70 = 30.

Therefore, the first option is NOT TRUE.

To check if option 2 is correct, calculate the lower limit to see if 70 is below the lower limit. If 70 is below the lower limit, then it is an outlier in the trail mix data.

Thus, Lower Limit = Q_1 - 1.5(IQR)

Q1 = 90,

IQR = 105 - 90 = 15

Lower Limit = 90 - 1.5(15)

Lower Limit = 90 - 22.5 = 67.5

70 is not less than the lower limit, therefore, 70 is not an outlier for the trail mix data. The second option is NOT TRUE.

The upper quartile of the trail mix data = 105.

The maximum value of the cracker data = 100.

Therefore, the third option is NOT TRUE.

Range can be used to determine how much variable there is in a data represented on a box plot. The greater the range value, the greater the variation.

Range of trail mix data = 115 - 70 = 45

Range of cracker data = 100 - 70 = 30.

<em>The range value for the number of calories in trail mix is greater than that for cracker, therefore, the number of calories in the packs of trail mix have a greater variation than the number of calories in the packs</em>

<em>of crackers.</em>

The fourth option is TRUE.

finlep [7]3 years ago
6 0

Answer: D. The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs of crackers.

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2 years ago
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A national survey conducted in 2011 among a simple random sample of 1,507 adults shows that 56% of Americans think the Civil War
Drupady [299]

Answer:

a)z=4.658

p_v =P(z>4.65)=1-P(z  

b) Using the significance level assumed \alpha=0.05 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life is  higher than 50%.

c) The 90% confidence interval would be given (0.527;0.593).

We are confident that about 54% to 59% of all Americans think the Civil War is relevant.

Step-by-step explanation:

I )Part a

1) Data given and notation

n=1507 represent the random sample taken  

X represent the Americans who thinks that the Civil War is still relevant to American politics and political life

\hat p estimated proportion of Americans who thinks that the Civil War is still relevant to American politics and political life in the sample

p_o=0.5 is the value that we want to test since the problem says majority    

\alpha=0.05 represent the significance level (no given, but is assumed)  

z would represent the statistic (variable of interest)  

p_v represent the p value (variable of interest)  

p= population proportion of Americans who thinks that the Civil War is still relevant to American politics and political life

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life exceeds 50%(Majority). :  

Null Hypothesis: p \leq 0.5

Alternative Hypothesis: p >0.5

We assume that the proportion follows a normal distribution.  

This is a one tail upper test for the proportion of  union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion \hat p is significantly (different,higher or less) from a hypothesized value p_o".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

np_o =1507*0.5=753.5>10

n(1-p_o)=1507*(1-0.5)=753.5>10

3) Calculate the statistic  

The statistic is calculated with the following formula:

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

On this case the value of p_o=0.5 is the value that we are testing and n = 1507.

z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1507}}}=4.658

The p value for the test would be:

p_v =P(z>4.65)=1-P(z  

II) Part b

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.  

Based on the alternative hypothesis the p value would be given by:

p_v =P(z>4.65)=1-P(z  

Using the significance level assumed \alpha=0.05 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life is  higher than 50%.

III) Part c

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.56 - 2.58 \sqrt{\frac{0.56(1-0.56)}{1507}}=0.527

0.56 + 2.58 \sqrt{\frac{0.56(1-0.56)}{1507}}=0.593

And the 90% confidence interval would be given (0.527;0.593).

We are confident that about 54% to 59% of all Americans think the Civil War is relevant.

And this result agrees with the result of part b, since the interval not contains the value of 0.5 we can conclude that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life it's higher than 0.5 at 90% of confidence.

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Which one gives you more time for your money?

We have to find the time per ticket, which is the number of minutes divided by the number of tickets.

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6/8 = 0.(60/8) = 0.75 minutes per ticket.

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