Question

USUULUYP43121bUOY-CV-WaUbben My OKstate Canvas b Billighway Geriatric Book UCO ASL Book 1 Spring 2020 Stat 2013/2053 - CRN 23116 Alfon nnect Elementary Statistics: A Step-By-Stop Approach, 10 Ed. MATH Hosted by ALEKS Corp Chapter 7 « Previous 11 12 13 14 15 16 17 18 Question 12 of 23 (3 points) 7.2 Section Exercise 10 The number of students who belong to the dance company at each of several randomly selected small universities is shown below. Round the sample statistics and final answers to one decimal place. 32 30 29 28 27 26 25 22 21 47 40 35 22 27 30 26 Download data Estimate the true population mean size of a university dance company with 98% confidence. Assume the variable is normally distributed. MacBook Air

Answer 1

Solution:

x	x2
32	1024
30	900
29	841
28	784
27	729
26	676
25	625
22	484
21	441
47	2209
40	1600
35	1225
22	484
27	729
30	900
26	676
40	1600
$\sum$x=507	$\sum$x2=15927

The sample mean is $\bar x$

Mean $\bar x$ = ($\sum$x / n) )

=32+30+29+28+27+26+25+22+21+47+40+35+22+27+30+26+40/17

=507/17

=29.8235

The sample standard is S

  S =$\sqrt{}$( $\sum$ x2 ) - (( $\sum$ x)2 / n ) n -1

=$\sqrt{}$15927-(507)217/16

=$\sqrt{}$15927-15120.5294/16

=$\sqrt{}$806.4706/16

=$\sqrt{}$50.4044

=7.0996

Degrees of freedom = df = n - 1 = 17 - 1 = 16

At 98% confidence level the t is ,

$\alpha$ = 1 - 98% = 1 - 0.98 = 0.02

$\alpha$ / 2 = 0.02 / 2 = 0.01

t$\alpha$ /2,df = t0.01,16 =2.583

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

= 2.583* (7.10 / $\sqrt$ 17)

= 4.45

Margin of error = 4.45

The 98% confidence interval estimate of the population mean is,

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

29.82 - 4.45 < $\mu$ < 29.82 + 4.45

25.37 < $\mu$ < 34.27

(25.37, 34.27 )