Statistics Problems
Question 6
The hypothesis tests whether the cost of the brunch is independent of the restaurant’s quality. The initial step is to formulate the null and alternative hypothesis.
Null hypothesis: Cost and quality are independent
Alternative hypothesis: Cost and quality are dependent
Chi-square statistic (X2) tests for the independence between variables
X2= ∑[(Oi -Ei)2/ Ei]
Oi and Ei are observed and expected frequency counts of the variables Don't use plagiarised sources.Get your custom essay just from $11/page
S
$.01-15.00
$15.01-30
$30.01+
Column Total
T
4
25
22
38
85
A
3
17
28
35
80
R
2
40
30
25
95
S
1
43
45
22
110
Row Total
125
125
120
370
E1,1= 85*125/370= 28.71 E1,2= 85*125/370= 28.71 E1,3=85*120/370= 27.57
E2,1= 80*125/370= 27.03 E2,2= 80*125/370= 27.03 E2,3= 80*120/370= 25.95
E3,1= 95*125/370= 32.09 E3,2= 95*125/370= 32.09 E3,3 = 95*120/370= 30.81
E4,1= 110*125/370= 37.16 E4,2= 110*125/370= 37.16 E4,3= 110*120/370= 35.68
Oi
Ei
(Oi- Ei)2/ Ei
25
28.7
0.477
22
28.7
1.564
38
27.6
3.919
17
27.0
3.704
28
27.0
0.037
35
26.0
3.116
40
32.1
1.944
30
32.1
0.137
25
30.8
1.092
43
37.2
0.904
45
37.2
1.635
22
35.7
5.257
X2= ∑[(Oi -Ei)2/ Ei]
23.786
The degrees of freedom are (r-1)(c-1)= (4-1)(3-1)= 6
The critical value at 6 degrees of freedom and a significance level of 0.01 is 16.81
Since chi-statistic is greater than the critical value (23.786>16.81), the correct decision is to reject the null hypothesis. Therefore, the cost of the brunch and the quality of the restaurant are dependent.
Question 11
The population mean is not given. By taking an arbitrary value of 48, the sample mean becomes 52. Given the eight observations, the standard deviation is 10.23
Null hypothesis: The population mean is equal to 48
Alternative hypothesis: The population mean is not equal to 48
The two-tailed t-test of the unpaired data of the sample from the normal distribution table is ±2.365. The calculated t-value is obtained from 52-48/(10.23/√8))= 1.106
At a 0.05 significance level, the p-value (0.3051) is less than the critical value. The null hypothesis cannot be rejected. Thus, there is insufficient evidence to conclude that all the signed acts will have the same probability of falling in each category of the eight groups.