Question #204349

These bags have a mean weight of 10kg with a standard deviation of 2kg.

a) Give the point estimate for the mean weight of all bags of cement in the warehouse. [ 1 marks ]

b) Compute the standard error. [1 marks ]

c) Construct the confidence interval of the mean weight of all the bags of cement in the warehouse

using 90% confidence level. [ 4 marks ]

d) If the inspector needs to estimate the mean weight of the bags with a maximum error of 0.5kg at the 90% confidence level, how large a sample will he need to take


1
Expert's answer
2021-06-08T17:32:45-0400

a) The sample mean xˉ\bar{x} is a point estimate of the population mean, μ.\mu.  

 The point estimate for the mean weight of all bags of cement in the warehouse is 10 kg.


b) SExˉ=s/nSE_{\bar{x}}=s/\sqrt{n}

Suppose we take 40 bags of cement in the warehous. Then the standard error is


SExˉ=sn=2400.3162SE_{\bar{x}}=\dfrac{s}{\sqrt{n}}=\dfrac{2}{\sqrt{40}}\approx0.3162

c) Suppose we take 40 bags of cement in the warehous.

The critical value for α=0.1\alpha=0.1 and df=n1=401=39df=n-1=40-1=39 degrees of freedom is tc=1.684875.t_c=1.684875.

The corresponding confidence interval is computed as shown below:


CI=(xˉtc×sn,xˉ+tc×sn)CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}},\bar{x}+t_c\times\dfrac{s}{\sqrt{n}})

=(101.684875×240,10+1.684875×240)=(10-1.684875\times\dfrac{2}{\sqrt{40}},10+1.684875\times\dfrac{2}{\sqrt{40}})

(9.4672,10.5328)\approx(9.4672, 10.5328)

Therefore, based on the data provided, the 90% confidence interval for the population mean is 9.4672<μ<10.5328,9.4672<\mu<10.5328, which indicates that we are 90% confident that the true population mean μ\mu is contained by the interval (9.4672,10.5328).(9.4672, 10.5328).


d) α=0.1,s=2\alpha=0.1, s=2


z1α/2;n1×sn0.5z_{1-\alpha/2;n-1}\times\dfrac{s}{\sqrt{n}}\leq0.5

n(2z1α/2;n1×s)2n\geq(2z_{1-\alpha/2;n-1}\times s)^2

n(4z1α/2;n1)2n\geq(4z_{1-\alpha/2;n-1})^2

n=45n=45


45(41.68023)2,False45\geq(4\cdot1.68023)^2, False

n=46n=46


46(41.679427)2,True46\geq(4\cdot1.679427)^2, True

He will need to take 46 bags.




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