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	nonconforming rooms
1	100	4
2	100	2
3	100	0
4	100	1
5	100	3
6	100	5
7	100	4
8	100	7
9	100	2
10	100	5
11	100	1
12	100	2
13	100	1
14	100	3
15	100	5
16	100	2
17	100	1
18	100	2
19	100	6
20	100	3
21	100	3
22	100	4
23	100	2
24	100	0
25	100	2

Whats p- value?

Conclusion?

Answer 1

Here we need to find proportion of non confronting rooms are less than 2% or not.As the null hypothesis always has an equal sign here

\

Here total percent of observed hotel romms/number of hotels = 2.8%=0.028

Here q=1-p=0.98

Now we will find test statistic

Now P-value for P(z<0.286)=0.6126

As the P-value(0.286) is greater than alpha=0.05 we fail to reject the null hypothesis.

So we do not have enough evidence to support the claim that the proportion of nonconforming rooms in less than 2​%.