Question

A statistics practitioner took a random sample of 54 observations from a population whose standard deviation is 33 and computed the sample mean to be 105. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 90% confidence. Confidence interval = C. Estimate the population mean with 99% confidence. Confidence Interval = Note: You can earn partial credit on this nenhlom

Answer 1

Solution:

a)

Degrees of freedom = df = n - 1 = 53

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,53 = 2.006

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

= 2.006 * (33 / $\sqrt$ 54)

= 9.008

The 95% confidence interval estimate of the population mean is,

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

105 - 9.008 < $\mu$ < 105 + 9.008

95.992 < $\mu$ < 114.008

(95.992 , 114.008)

b)

At 90% confidence level the t is ,

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

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

t_{$\alpha$ /2,df} = t_0.05,53 = 1.674

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

= 1.674 * (33 / $\sqrt$ 54)

= 7.517

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

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

105 - 7.517 < $\mu$ < 105 + 7.517

97.483 < $\mu$ < 112.517

(97.483 , 112.517)

c)

At 99% confidence level the z is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

t_{$\alpha$ /2,df} = t_0.005,53 = 2.672

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

= 2.672 * (33 / $\sqrt$ 54)

= 11.999

The 99% confidence interval estimate of the population mean is,

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

105 - 11.999 < $\mu$ < 105 + 11.999

93.001 < $\mu$ < 116.999

(93.001 , 116.999)