Answer to Question #204349 in Statistics and Probability for Benjamin ocran

Question

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

Answer 1

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

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

b) $SE_{\bar{x}}=s/\sqrt{n}$

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

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

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

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

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}},\bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

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

$$\approx(9.4672, 10.5328)$$

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

d) $\alpha=0.1, s=2$

$$z_{1-\alpha/2;n-1}\times\dfrac{s}{\sqrt{n}}\leq0.5$$

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

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

$n=45$

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

$n=46$

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

He will need to take 46 bags.