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Ugo [173]
3 years ago
10

What categories are the same percentage of the budget? Category Amount spent per month Housing $1,302 Taxes $1,550 Food $558 Clo

thing $124 Entertainment $124 Transportation $372 College Savings $558 Retirement Savings $558 Gasoline $248 Emergency Savings $310 Utilities $496
Mathematics
2 answers:
valina [46]3 years ago
5 0

To be the same percentage, the numbers have to be the same.

Clothing is $124 and Entertainment is $124, so these would be the same percentage.

Food is $558, College savings is $558 and Retirement Savings is $558, these are identical so would also be the same percentage.

xxMikexx [17]3 years ago
4 0
I believe the answer is B, there is a lot going on tbh but trust me, it’s b fam
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Answer:

t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626    

The degrees of freedom are given by:

df=n-1=8-1=7  

The p value would be given by:

p_v =P(t_{(7)}  

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

Step-by-step explanation:

Information given

60, 56, 60, 55, 70, 55, 60, and 55.

We can calculate the mean and deviation with these formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

Replacing we got:

\bar X=58.875 represent the mean

s=5.083 represent the sample standard deviation for the sample  

n=8 sample size  

\mu_o =60 represent the value that we want to test

\alpha=0.1 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is less than 60, the system of hypothesis would be:  

Null hypothesis:\mu \geq 60  

Alternative hypothesis:\mu < 60  

The statistic would be given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info we got:

t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626    

The degrees of freedom are given by:

df=n-1=8-1=7  

The p value would be given by:

p_v =P(t_{(7)}  

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

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