Question

Use the confidence level and sample data to find a confidence interval for estimating the population u. Round your answer to the same number of decimal places as the sample mean. 20) A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 20) 5.2 on a placement test. What is the 90% confidence interval for the mean score,, of all students taking the test? A) 27.8 <u <31. 2 B ) 28.2<u<30. 8 C ) 28.4 <u<30.6 D) 27.9<<31.1

Answer 1

Solution :

Given that,

$\bar x$ = 29.5

s = 5.2

n = 59

Degrees of freedom = df = n - 1 = 59 - 1 = 58

At 90% confidence level the z is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.10

$\alpha$ / 2 = 0.10 / 2 = 0.05

t $\alpha$ /2,df = t0.05,58 =1.672

Margin of error = E = t $\alpha$ /2,df * (s / $\sqrt$ n)

= 1.672 * (5.2 / $\sqrt$ 59)

= 1.1

Margin of error = 1.1

The 90% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

29.5 - 1.1 < $\mu$ < 29.5 + 1.1

28.4 < $\mu$ < 30.6

(28.4,30.6 )

Option C ) is correct.