Question

20 people are randomly chosen at a shopping mall to taste-test a new brand of fruit drink. They are asked to rate the drink on a scale from 1 to 7, with 1 being very bad and 7 being very good. The results of the survey reveal that the average rating is 3.50 with a standard deviation of 2.00. The marketing division of the fruit drink distributor is only interested in selling this drink if the true mean rating is more than 3.80. Use the significance level of 0.05.

Answer 1

Answer:

a)

Given,

sample n = 20

x = 3.50

standard deviation = 2

population mean u = 3.80

significance level = 0.05

Null hypothesis Ho : u = 3.80

Alternative hypothesis Ha : u > 3.80

b)

consider,

test statistic = (x - u)/(s/sqrt(n))

substitute values

= (3.50 - 3.80)/(2/sqrt(20))

test statistic = -0.67

Option C is right answer.

c)

Here degree of freedom = n-1 = 20-1 = 19

t(alpha/2 , df) = t(0.025 , 19) = 1.729

P value = 0.25546 [since from t table]

i.e.,

Option A is right answer.

d)

Here 0.255 < 0.5 , so we fail to reject Ho.

So there is no sufficient evidence.