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Nastasia [14]
3 years ago
8

On Monday, it took 5 builders 3 and a 1/2 hours to build a wall. An identical wall needs to be built on Tuesday but only 2 build

ers are available. Each builder is paid £9.30 for each hour or part of an hour they work. Work out how much each builder will be paid for the work completed by Tuesday.
Mathematics
1 answer:
alexira [117]3 years ago
6 0
5 \text { builders =} 3 \dfrac{1}{2}  \text { hours }

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Find 1 worker :
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1 \text { builders =} 3 \dfrac{1}{2}  \times 5 \text { hours }

1 \text { builders =} \dfrac{35}{2} \text { hours }

-------------------------------------------------------
Find 2 workers :
-------------------------------------------------------
2 \text { builders =} \dfrac{35}{2} \div 2 \text { hours }

2 \text { builders =} \dfrac{35}{4} \text { hours }

2 \text { builders =} 8.75 \text { hours }

-------------------------------------------------------
Find cost of 1 worker :
-------------------------------------------------------
Since they are paid hourly or part of an hour at £9.30.

One builder = £9.30 x 9 = £83.70

-------------------------------------------------------
Answer: Each worker will be paid £83.70.
-------------------------------------------------------


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p_v =P(t_{(18)}

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Step-by-step explanation:

Data given and notation

We can calculate the sample 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}}

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\bar X_{2}=23.3 represent the mean for the sample mean for 2

s_{1}=4.818 represent the sample standard deviation for the sample 1

s_{2}=5.559 represent the sample standard deviation for the sample 2

n_{1}=10 sample size selected 1

n_{2}=10 sample size selected 2

\alpha represent the significance level for the hypothesis test.

t 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.

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Null hypothesis:\mu_{1} \geq \mu_{2}

Alternative hypothesis:\mu_{1} < \mu_{2}

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

P-value

The first step is calculate the degrees of freedom, on this case:

df=n_{1}+n_{2}-2=10+10-2=18

Since is a one sided test the p value would be:

p_v =P(t_{(18)}

Conclusion

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

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