Answer to Question #335413 in Statistics and Probability for loe

Question

1. Miss Romero noted that the mean scores of a random sample of 15 Grade 8 students who had taken a special test were 80.5. If the standard deviation of the scores was 3.1, and the sample came from an approximately normal population, find the point and the interval estimates of the population mean, using 95%confidence.

2.    Edith has observed that the mean age of 18 youth volunteers in a barangay project is 17.5 years with a standard deviation of 2years. If the sample comes from an approximately normal distribution, what are the point and the interval estimates of the population mean? Adopt a 99% confidence level.

 

 

 

 

Answer 1

1. Point estimate $\bar{x}=80.5.$

The critical value for $\alpha = 0.05,df=n-1=7$ degrees of freedom is

is $t_c = z_{1-\alpha/2; n-1} = 2.364619.$

The corresponding confidence interval is computed as shown below:

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

$$=(80.5-2.364619\times\dfrac{3.1}{\sqrt{8}}, 80.5+2.364619\times\dfrac{3.1}{\sqrt{8}})$$

$$=(77.908, 83.092)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $77.908 < \mu < 83.092,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(77.908, 83.092) .$

2. Point estimate $\bar{x}=17.5.$

The critical value for $\alpha = 0.01,df=n-1=17$ degrees of freedom is

is $t_c = z_{1-\alpha/2; n-1} =2.89823.$

The corresponding confidence interval is computed as shown below:

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

$$=(17.5-2.89823\times\dfrac{2}{\sqrt{18}}, 17.5+2.89823\times\dfrac{2}{\sqrt{18}})$$

$$=(16.134, 18.866)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $16.134 < \mu < 18.866,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(16.134, 18.866) .$