Question

In: Statistics and Probability

Construct a 95% confidence interval to estimate the population mean of printable page numbers from the new printer and PLEASE interpret the result.

Number of Pages Printed
2120 1953 2289 1897 2187 2283 2321 1956 2551 2423 2034 2221 2507 2110 2043 2476 2067 2119 1889 1734 2501 2333 2139 1998

Construct a 95% confidence interval to estimate the population mean of printable page numbers from the new printer and PLEASE interpret the result.

d) Repeat the same questions as (c) with 90% confidence level

Expert Solution

Solution:

x	x2
2120	4494400
1953	3814209
2289	5239521
1897	3598609
2187	4782969
2283	5212089
2321	5387041
1956	3825936
2551	6507601
2423	5870929
2034	4137156
2221	4932841
2507	6285049
2110	4452100
2043	4173849
2476	6130576
2067	4272489
2119	4490161
1889	3568321
1734	3006756
2501	6255001
2333	5442889
2139	4575321
1998	3992004
∑x=52151	∑x2=114447817

a ) Mean ˉx=∑xn

=2120+1953+2289+1897+2187+2283+2321+1956+2551+2423+2034+2221+2507+2110+2043+2476+2067+2119+1889+1734+2501+2333+2139+1998/24

=52151/24

=2172.9583

b ) Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√114447817-(52151)22423

=√114447817-113321950.041723

=√1125866.958323

=√48950.7373

=221.2481

Degrees of freedom = df = n - 1 = 24- 1 = 23

c ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t_/2,df = t_0.025,23 =2.069

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

= 2.069 * (221.2 / $\sqrt$ 24)

= 93.4

Margin of error = 93.4

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

- E < < + E

2172.9 - 93.4 < < 2172.9 + 93.4

2079.5 < < 2266.3

(2079.5, 2266.3 )

d ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t_/2,df = t_0.05,23 = 1.714

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

= 1.714 * (221.2 / $\sqrt$ 24)

= 77.3

Margin of error = 77.3

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

- E < < + E

2172.9 - 77.3 < < 2172.9 + 77.3

2095.5 < < 2250.3

(2079.5, 2266.3 )

orchestra answered 2 years ago

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