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aleksley [76]
3 years ago
7

Su Ling started an art project with 1 yard felt. She used 5/6 yard. How much felt does Su Ling have left?

Mathematics
2 answers:
Liula [17]3 years ago
8 0
She would have 1/6 or 0.5 feet
Vadim26 [7]3 years ago
7 0
Su has about 1/6 yards of felt left my dude/dudett
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Null hypothesis:\mu = 100  

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Step-by-step explanation:

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Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

We can calculate the sample mean and deviation for this data with the following 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}}

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\bar X=98.375 represent the sample mean  

s=0.6.109 represent the sample standard deviation  

n=12 sample size  

\mu_o =100 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :  

Null hypothesis:\mu = 100  

Alternative hypothesis:\mu \neq 100  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921  

4) P-value  

First we need to find the degrees of freedom for the statistic given by:

df=n-1=12-1=11

Since is a two sided test the p value would given by:  

p_v =2*P(t_{11}  

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If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.  

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