Question

A simple random sample of 64 concert tickets was drawn from a normal population. The mean and standard deviation of the sample were $120 and $25, respectively.

Estimate the population mean with 90% confidence.

	a.	LCL =115.95 UCL = 124.05
	b.	LCL =114.85 UCL = 125.14
	c.	LCL =115.99 UCL = 124.01
	d.	LCL =114.78 UCL = 125.22

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = $120

sample standard deviation = s = $25

sample size = n = 64

Degrees of freedom = df = n - 1 = 64-1= 63

At 90% confidence level the t is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.1

$\alpha$ / 2 = 0.1/ 2 = 0.05

t_{$\alpha$ /2,df} = t_0.05,63 = 1.669

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

= 1.669 * (25 / $\sqrt$ 64)

= 5.22

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

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

120 - 5.22 < $\mu$ < 120 + 5.22

114.78 < $\mu$ < 125.22

(114.78,125.22)

LCL = 114.78

UCL = 125.22

Option d is correct.