Question

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Question 3 (4 marks) Part a) Given the following information X-500, σ=12, n=50 i) Determine the 95% confidence interval estimate of population mean. ii) Determine the 99% o confidence interval estimate of population mean Part b) A statistics practitioner calculated the mean and standard deviation from a sample of 51. They are X-120 and s-15. (i) Estimate the population mean with 95% confidence level (ii) Estimate the population mean with 99% confidence level.

Answer 1

Answer:

Given that:

Part (a) (i)   $\dpi{100} \bar{X} = 500$ , $\dpi{100} \sigma =12$ , n= 50

We know that

The confident level is 95%.

Here $\dpi{100} \bar{X} - E < \mu < \bar{X} &plus; E$

We consider

$\dpi{100} E = Z_{c} \times \frac{r}{\sqrt{n}}$

e=95% ,   $\dpi{100} \alpha = 0.05$

Using critical value distribution   $\dpi{100} Z_{c} =1.96$

$\dpi{100} E = Z_{c} \times \frac{r}{\sqrt{n}}$

  $\dpi{100} E=1.96\times \frac{12}{\sqrt{50}}$

= $\dpi{100} 1.96\times 1.69705$

   =  $\dpi{100} 3.3262$

  $\dpi{100} \bar{X}-E=500-3.3262=496.6738.$

  $\dpi{100} \bar{X&plus;} E= 500&plus;3.3262=503.3262$.

The confident level is (496.6738 ,503.3262)

(ii) To find the 99% of confidence level

We know that  $\dpi{100} \alpha =0.01$$\dpi{100} Z_{c}=2.575$

  

$\dpi{100} E = Z_{c} \times \frac{r}{\sqrt{n}}$

$\dpi{100} E=2.575\times \frac{12}{\sqrt{50}}$

$\dpi{100} E=2.575\times 1.6970$

  $\dpi{100} E=4.3699$

$\dpi{100} \bar{X}-E=500-4.3699=495.6301$
  $\dpi{100} \bar{X}&plus; E=500&plus;4.3699=504.3699$

The 99% of confidence level is (495.6301 ,504.3699)

Part (b) (i) We know that $\dpi{100} \bar{X}=120 , S=15 , n= 51$

95% of confident interval to estimated

Here $\dpi{100} \bar{X} - E < \mu < \bar{X} &plus; E$

$\dpi{100} E=t_{c} \times \frac{S}{\sqrt{n}}$

Degree of freedom = 51-1=50 , $\dpi{100} \alpha =0.05$

Critical value for t-distribution $\dpi{100} t_{c}=2.00g$

   $\dpi{100} E=t_{c} \times \frac{S}{\sqrt{n}}$

  $\dpi{100} E=2.00\times \frac{15}{\sqrt{51}}$

$\dpi{100} =2.00\times 2.1004$

$\dpi{100} E=4.2197$
$\dpi{100} \bar{X}-E=120.4.2197=115.7803$

  $\dpi{100} \bar{X}&plus;E=120&plus;4.2197=124.2197$

95% of confident level is (115.7803,124.2197)

(ii) 99% of confident level

$\dpi{100} \alpha =0.01 , t_{c}= 2.678$

     $\dpi{100} E=t_{c} \times \frac{S}{\sqrt{n}}$

$\dpi{100} E=2.678\times \frac{15}{\sqrt{51}}$

  $\dpi{100} =2.678\ \times 2.1004$

  $\dpi{100} E=5.6249$

$\dpi{100} \bar{X}-E=120-5.6249=114.3751$   

$\dpi{100} \bar{X}&plus;E=120&plus;5.6249=125.6249$

The 99% confident interval is (114.3751,125.6249)