Question

Bhavna is assessing the possibility of opening an eatery in Suva. The Fiji Chamber of Commerce evaluates that 'Bhavna families, on average, eat out is at least 3 nights of every week'. Singh intends to test this hypothesis at the 0.01 level of significance. His random sample of 81 Bhavna families delivered a mean of 2.7. Assuming that the population standard deviation is 0.9 evenings per week, the appropriate decision is

Answer 1

Solution :

$\mu$= 3

= 2.7

T

$\sigma$ = 0.9

n = 81

This is the right tailed test .

The null and alternative hypothesis is ,

H0 :  $\mu$  = 3

Ha : $\mu$   $\geq$ 3

Test statistic = z

= (
T
- $\mu$ ) / $\sigma$ / $\sqrt$ n

= (2.7-3) / 0.9 / $\sqrt$ 81

= -3

Test statistic = z = -3

P(z > -3 ) = 1 - P(z < -3 ) = 1 - 0.0013

P-value =0.9987

$\alpha$ = 0.01

P-value > $\alpha$

0.9987 > 0.01

Fail to reject the null hypothesis .

There is insufficient evidence to suggest that