Question

Use SPSS to get the appropriate output, and please explain the process. Along with providing the SPSS output, use the 5 steps of hypothesis testing to analyze the results by using the p-value approach. 6. Create an SPSS file with the variable names Hotel and Inspection. In the Hotel column, enter the value 1 in the first 25 cases and the value 2 in the next 30 cases. Then, in the Inspection column, enter the value 1 for any four cases of Hotel 1 and 0 for the rest of the Hotel 1 cases. And, enter the value 1 for any 10 cases of Hotel 2 and 0 for the rest of the Hotel 2 cases. For the inspection variable, 1 = a room failed inspection & 0 = a room passed inspection. a) Run a binomial test on for H0: π≤ 0.1 versus H1: π > 0.1 Test at 0.05 level of significance. b) Run a chi‐square test on H0: π1 – π2 = 0 versus H1: π1 – π2 ≠ 0 Test at 0.05 level of significance.

Answer 1

Use SPSS to get the appropriate output, and please explain the process. Along with providing the SPSS output, use the 5 steps of hypothesis testing to analyze the results by using the p-value approach. 6. Create an SPSS file with the variable names Hotel and Inspection. In the Hotel column, enter the value 1 in the first 25 cases and the value 2 in the next 30 cases. Then, in the Inspection column, enter the value 1 for any four cases of Hotel 1 and 0 for the rest of the Hotel 1 cases. And, enter the value 1 for any 10 cases of Hotel 2 and 0 for the rest of the Hotel 2 cases. For the inspection variable, 1 = a room failed inspection & 0 = a room passed inspection.

a) Run a binomial test on for H0: π≤ 0.1 versus H1: π > 0.1 Test at 0.05 level of significance.

binomial test for inspection.

SPSS commond:

Analyze---Nonparametric tests---Legacy Dialogs---Binomial

Select Hotel in the test variable list and enter Test proportion as 0.10

NPAR TESTS

/BINOMIAL (0.10)=Inspection

/MISSING ANALYSIS.

SPSS output

Binomial Test
		Category	N	Observed Prop.	Test Prop.	Exact Sig. (1-tailed)
Inspection	Group 1	a room failed inspection	14	.3	.1	.001
	Group 2	a room passed inspection	41	.7
	Total		55	1.0

Calculated P=0.001 which is < 0.05 level of significance.

Ho is rejected.

We conclude that population proportion is >0.10.

b) Run a chi‐square test on H0: π1 – π2 = 0 versus H1: π1 – π2 ≠ 0 Test at 0.05 level of significance.

SPSS syntax:

Analyze--- Descriptive statistics--- Crosstabs-----select Htel in Rows box and Inspection in the column box

CROSSTABS

/TABLES=Hotel BY Inspection

/FORMAT=AVALUE TABLES

/STATISTICS=CHISQ

/CELLS=COUNT

/COUNT ROUND CELL.

SPSS output:

Hotel * Inspection Crosstabulation
Count
		Inspection		Total
		a room passed inspection	a room failed inspection
Hotel	1	21	4	25
	2	20	10	30
Total		41	14	55

Chi-Square Tests
	Value	df	Asymptotic Significance (2-sided)	Exact Sig. (2-sided)	Exact Sig. (1-sided)
Pearson Chi-Square	2.159a	1	.142
Continuity Correctionb	1.342	1	.247
Likelihood Ratio	2.226	1	.136
Fisher's Exact Test				.215	.123
Linear-by-Linear Association	2.120	1	.145
N of Valid Cases	55
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 6.36.
b. Computed only for a 2x2 table

Calculated chi square =2.159, P=0.142 which is > 0.05 level of significance.

Ho is not rejected.

There is not sufficient evidence to conclude that the two population proportions are different.