Question

A hair dresser placed a $400 advertisement in the local news paper to increase her daily revenue. The table below shows revenues before and after the advertisement.

Day	Before advertainment	After advertisement	d̅ (After -Before)
Monday	500	550	50
Tuesday	485	525	40
Wednesday	440	570	130
Thursday	450	620	170
Friday	540	550	10

Assume that H0: µd ≤0 HA: µd >0. d̅ = After – Before. Answer the following questions.

The mean of d̅ is:

75

70

80

85

The standard deviation of d̅ is

66.08

67.08

68.08

69.08

To test the hypothesis, we use:

t-test

z-test

both z and t-tests

m-test

The test statistic value is:

2.65

2.35

2.45

2.67

If the level of significance is 5 %, the critical value is:

2.138

2.132

2.12

2.15

The decision is:

Do not reject the null hypothesis because it is a right tail test

Do not reject the null hypothesis because it is a left tail test

Reject the null hypothesis

Reject the alternative hypothesis

The interpretation is:

There is no enough evidence to support the alternative hypothesis

There is no enough evidence to support the null hypothesis

There is enough evidence to support the null hypothesis

None of the above

Answer 1