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harkovskaia [24]
3 years ago
6

Helen practiced the piano 5 days last week. The list below shows the number of minutes she practiced each day: 20,25,30,20,30 Wh

at is the MAD (mean absolute deviation) in Helen's practice times?
Mathematics
1 answer:
LekaFEV [45]3 years ago
8 0

Answer:

4

Step-by-step explanation:

Given the data:

X : 20,25,30,20,30

Mean absolute deviation :

Σ(x - xbar) / n

n = sample size = 5

xbar = ΣX / n = (20+25+30+20+30) /5

xbar = 25

((20-25) + (25-25) + (30-25) + (20-25) + (30-25)) / 5

We take the absolute values, hence, negative signs are ignored :

(5 + 0 + 5 + 5 + 5) / 5

20 / 5

= 4

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Answer:

t=\frac{6.3-6.0}{\frac{1}{\sqrt{26}}}=1.530  

df = n-1 = 26-1=25

Since is a right-sided test the p value would be:  

p_v =P(t_{25}>1.530)=0.0693  

If we compare the p value and the significance level given \alpha=0.025 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is significantly higher than 6.0

Step-by-step explanation:

Data given and notation  

\bar X=6.3 represent the sample mean    

s=1.0 represent the sample standard deviation

n=26 sample size  

\mu_o =6.0 represent the value that we want to test  

\alpha=0.025 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the ture mena is higher than 6.0, the system of hypothesis would be:  

Null hypothesis:\mu \leq 6.0  

Alternative hypothesis:\mu > 6.0  

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".  

Calculate the statistic  

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

t=\frac{6.3-6.0}{\frac{1}{\sqrt{26}}}=1.530  

P-value  

The degrees of freedom are given by:

df = n-1 = 26-1=25

Since is a right-sided test the p value would be:  

p_v =P(t_{25}>1.530)=0.0693  

Conclusion  

If we compare the p value and the significance level given \alpha=0.025 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is significantly higher than 6.0

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Answer:

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