Question

We wish to give a 90% confidence interval for the mean value of a normally distributed random variable. We obtain a simple random sample of 9 elements from the population on which the variable is defined. The sample has a mean value of 10.2 and a sample standard deviation 1.5. Find the 90% confidence interval for the mean value.

Answer 1

Answer: $(9.27025,\ 11.12975)$

Step-by-step explanation:

Given : Sample size : n= 9

Degree of freedom = df =n-1 =8

Sample mean : $\overline{x}=10.2$

sample standard deviation : $s= 1.5$

Significance level ; $\alpha= 1-0.90=0.10$

Since population standard deviation is not given , so we use t- test.

Using t-distribution table , we have

Critical value = $t_{\alpha/2, df}=t_{0.05 , 8}=1.8595$

Confidence interval for the population mean :

$\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}$

90% confidence interval for the mean value will be :

$10.2\pm (1.8595)\dfrac{1.5}{\sqrt{9}}$

$10.2\pm (1.8595)\dfrac{1.5}{3}$

$10.2\pm (1.8595)(0.5)$

$10.2\pm (0.92975)$

$(10.2-0.92975,\ 10.2+0.92975)$

$(9.27025,\ 11.12975)$

Hence, the 90% confidence interval for the mean value= $(9.27025,\ 11.12975)$