Question

The accompanying data shows the weekly purchases of printers at a particular electronic store. Using

alpha α equals=0.050

perform a​ chi-square test to determine if the number of printers sold per week follows a normal probability distribution. Note that

x equals=11.2

and

s equals=4.5
Click the icon to view the weekly purchases of printers.

Observed weekly purchases of printers 8 14 11 18 11 15 4 7 8 3 16 16 14 9 12 18 12 13 5 10 7 10 10 10 18 8 17 5 19 10 4 8 12 6 14 11 13 13 14 11 11 18 20 Print Done X co

Use the intervals below to calculate the​ chi-square test​ statistic, χ2.

Interval​ 1:			z	≤−1.0
Interval​ 2:−1.0	<	z	≤	0
Interval​ 3:	0	<	z	≤	1.0
Interval​ 4:	1.0	<	z

chi squared χ2 = ?

​(Round to two decimal places as​ needed.)

Determine the​ p-value.

​p-value =??

​(Round to three decimal places as​ needed.)

"blank H0.There "Blank" enough evidence to conclude that the number of printers sold per week at the electronic store does not follow the normal probability distribution.

Answer 1

Let f_o and f_e be observed and expected frequencies.

Total number of observations, n = 45

Intervals	Area in Class	X = 11.2 + z * 4.5	fo	fe = Area * 45	(fo - fe)	(fo - fe)2	(fo - fe)2/fe
z $\le$ -1.0	0.1587	X $\le$ 6.7	6	7.1415	-1.1415	1.303022	0.1824577
-1.0 < z $\le$ 0	0.3413	6.7 < z $\le$ 11.2	18	15.3585	2.6415	6.977522	0.4543101
0 < z $\le$ 1.0	0.3413	11.2 < X $\le$ 15.7	11	15.3585	-4.3585	18.99652	1.236873
1.0 < z	0.1587	15.7 < X	10	7.1415	2.8585	8.171022	1.14416

Area in class are calculated from z tables -

P(z $\le$ -1.0) = 0.1587

P(-1.0 < z $\le$ 0) = P(z <0) - P(z $\le$ -1.0) = 0.5 - 0.1587 = 0.3413

P(0 < z $\le$ 1.0) = P(z <1) - P(z $\le$ 0) = 0.8413 - 0.5 = 0.3413

P(z > 1.0) = 0.1587

X = = 11.2 + z * 4.5

S

chi-square test statistic X^2  = Sum of (f_o - f_e)² /f_e = 0.1824577 + 0.4543101 + 1.236873 + 1.14416 = 3.017801

Degree of freedom = k - p - 1

where k is number of intervals

p is number of parameters. Here mean and standard deviation are the parameters. p = 2

Degree of freedom = 4 - 2 - 1= 1

P(X^2 > 3.017801, df = 1) = 0.082355

Thus, p-value is 0.082