Question

The accompanying table provides the data for 100 room inspections at each of 25 hotels in a major chain. Management would like the proportion of nonconforming rooms to be less than 2​%. Formulate a​ one-sample hypothesis test for a proportion and perform the calculations using the correct formulas and Excel functions. Use a level of significance of 0.05.

Sample	Rooms Inspected	Nonconforming Rooms	Fraction Nonconforming
1	100	4	0.04
2	100	1	0.01
3	100	0	0
4	100	1	0.01
5	100	3	0.03
6	100	6	0.06
7	100	4	0.04
8	100	7	0.07
9	100	2	0.02
10	100	5	0.05
11	100	1	0.01
12	100	2	0.02
13	100	2	0.02
14	100	4	0.04
15	100	5	0.05
16	100	2	0.02
17	100	1	0.01
18	100	2	0.02
19	100	6	0.06
20	100	2	0.02
21	100	4	0.04
22	100	5	0.05
23	100	1	0.01
24	100	0	0
25	100	1	0.01

Is there sufficient evidence at the 0.05 level of significance that the proportion of nonconforming rooms to be less than 2​%?

Determine the null​ hypothesis, H0​, and the alternative​ hypothesis, H1. Then compute the test statistic and p-value. Finally, state the conclusion.

Answer 1

Ho :   p ≥ 0.02  
H1 :   p < 0.02   (Right tail test)
          
Level of Significance,   α =    0.05  

Sample Size,   n =    100  
          
Sample Proportion ,    p̂ = 0.0284  
          
Standard Error ,    SE = √( p(1-p)/n ) = √(0.0284(1-0.0284)/100)= 0.01400  
Z Test Statistic = ( p̂-p)/SE =    (0.0284-0.02)/0.014=   0.60000  
          

          
p-Value   = 0.7257 [Excel function =NORMSDIST(-z)
Decision:   p value>α ,do not reject null hypothesis       

p value is greater than α, so, fail to reject the null

There is no enough evidence ......

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