Answer to Question #111172 in Statistics and Probability for pro

Question

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Answer 1

Assume the population distribution is normally distributed, the weights (in 1000 grams) of a random sample of 10 boxes of cereal are:

$\begin{matrix} 10.2 & 10.1 & 10.1 & 9.7 & 10.3 & 9.8 & 9.9 & 9.8 & 10.4 & 10.3 \end{matrix}$

(a) Find a 99% confidence interval for the mean weight of each box of cereal

$\displaystyle\sum_{i=1}^{10}x_i=10.2+10.1+10.1+9.7+10.3+9.8+$ $+9.9+9.8+10.4+10.3=100.6$

$\displaystyle\sum_{i=1}^{10}x_i^2 =10.2^2 +10.1^2 +10.1^2 +9.7^2+10.3^2+$ $+9.8^2+9.9^2+9.8^2+10.4^2+10.3^2=1012.58$ $\bar{X}={\displaystyle\sum_{i=1}^{10}x_i\over 10}={100.6 \over 10}=10.06$

$s^2={1 \over 10-1}\bigg(\displaystyle\sum_{i=1}^{10}x_i^2-{1 \over 10}\big(\displaystyle\sum_{i=1}^{10}x_i\big)^2\bigg)=$

$={1 \over 10-1}\bigg(1012.58-{1 \over 10}\big(100.6\big)^2\bigg)={0.544 \over 9}$

$s=\sqrt{s^2}=\sqrt{{0.544 \over 9}}\approx0.2459$

The critical value for $\alpha=0.01$ and $df=n-1=9$ degrees of freedom is $t_c=3.2498.$

The corresponding confidence interval is computed as shown below:

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

$=(10.06-3.2498 \times{0.2459 \over \sqrt{10}},10.06+3.2498 \times{0.2459 \over \sqrt{10}})\approx$

$\approx(9.8073,10.3127)$

Therefore, based on the data provided, the 99%c onfidence interval for the population mean is $9.8073<\mu<10.3127,$ which indicates that we are 99% confident that the true population mean $\mu$

is contained by the interval $(9.8073,10.3127).$

(b) Assume the population standard deviation is 30 grams. Construct a 96% confidence interval for the mean weight of each box of cereal.

We need to construct the 96% confidence interval for the population mean $\mu.$ The following information is provided: $\bar{X}=10.06, \sigma=0.03, n=10.$

The critical value for $\alpha=0.04$ is $z_c=z_{1-\alpha/2}=2.054.$ The corresponding confidence interval is computed as shown below:

$CI=(\bar{X}-z_c \times{\sigma \over \sqrt{n}},\bar{X}+z_c \times{\sigma \over \sqrt{n}})=$

$=(10.06-2.054 \times{0.03 \over \sqrt{10}},10.06+2.054 \times{0.03 \over \sqrt{10}})\approx$

$\approx(10.0405,10.0795)$

Therefore, based on the data provided, the 99%c onfidence interval for the population mean is 

$10.0405<\mu<10.0795,$ which indicates that we are 99% confident that the true population mean  $\mu$

is contained by the interval $(10.0405,10.0795).$

(c) Assume the population standard deviation is given. If one decreases the sample size while keeping the same confidence level, what is the effect of the length of a confidence interval of a population mean?

$length=2z_c \times{\sigma \over \sqrt{n}}$

The length of a confidence interval increases.

(d) Assume the population standard deviation is given. If one increases the confidence level while keeping the same sample size, what is the effect of the length of a confidence interval of a population mean?

Increasing the confidence level increases the error bound, making the confidence interval wider.

$length=2z_c \times{\sigma \over \sqrt{n}}$

The length of a confidence interval increases.