Question

A random sample of 200 tins of coconut oil gave an average weight of 4.95 kg. With a standard deviation of 0.21 kg. Do we accept that net weight is 5kg per tin at 5% los.

Answer 1

add kg then add 200 turn in it kg and divide it with 5

Answer 2

Answer: Accepted

Given: sample size(n)=200

Average weight ; mean (μ)= 4.95

Standard deviation (σ)= 0.21

Net weight (x̅ ) = 5

To Find: Acceptance of H0 = 5 ; at 5% level

Step-by-step explanation:

Here, it is a null hypothesis that H0=5

We know the formula for a null hypothesis

In |z| = | x̅  - μ/ σ √n |

If |z| < 1.96 the difference is not significant at the 5% level and H0 is accepted, otherwise rejected.

Now inserting the given values in the above formula

|z|= | 5-4.95/0.21√200|

   = 0.016< 1.96

Here, the value is less than 1.96, thus the sample is accepted at 5% level.

Project code: #SPJ2