Question

A brand of chocolate bar has a stated weight of 6 oz. with s= 0.25 oz. A sample of 9 bars has an average weight of 6.05 oz.

Test H₀: µ = 6 oz. H₁: µ ≠ 6 oz. at the 5% significance level.

Answer 1

Solution :

This is the two tailed test,  

The null and alternative hypothesis is ,

H₀ :   = 6

H₁ :    6

Test statistic = t =

= ( - ) / s / n

= (6.05 - 6 ) / 0.25 / 9

Test statistic = t = 0.60

degrees of freedom = n - 1 = 9 - 1 = 8

P(t > 0.60) = 1-P (t < 0.60) = 1 - 0.7174 = 0.2826

P-value = 2 * P(t > 0.60)

P-value = 2 * 0.2826

P-value = 0.5652

= 0.05  

P-value >

Fail to reject the null hypothesis