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Statistics

Statistics Problems

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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

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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.

 

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