Answer to Question #160722 in Statistics and Probability for unknown

Question

The following are prices of a sanitizer sold at a few selected stores in a District A during the COVID-19 pandemic. Based on the sample data, at 0.05 level of significance, can we conclude the average price of the sanitizer in District A is RM25? Make your conclusion using the p-value approach, with the aid of Mic. Office EXCEL.

RM20.00, RM20.50, RM26.00, RM29.50, RM20.70, RM20.20, RM18.70, RM23.00 

Answer 1

H_0: μ = 25

H_1: μ ≠ 25

Sample mean:

$\bar{X}=AVERAGE(A1:A8) \\ = 22.325$

Standard deviation:

S =STDEV(A1:A8)

= 3.67

The sample size:

n =COUNT(A1:A8)

= 8

$t = \frac{\bar{X}-μ}{S/ \sqrt{n}} \\ = \frac{22.32 - 25}{3.67 / \sqrt{8}} \\ = \frac{-2.68}{1.30} \\ = -2.06$

α = 0.05

df = 8 -1 = 7

$t_{crit}$ = TINV(2 * α, df)

= TINV(0.1, 7)

= 1.89

|t| = 2.06 > t_{crit} = 1.89

We can conclude that the null hypothesis is rejected.

Using the P-value approach:

p-value = TDIST(t, df, 1)

= TDIST(2.06, 7, 1)

= 0.039

Since p = 0.039 < 0.05

We can conclude that the null hypothesis is rejected.

We conclude the average price of the sanitizer in District A is not RM25.