In: Statistics and Probability
5. An analysis of the fat content, X % of a random sample of 175 particular burgers resulted in the following table.
Fat Content | No of Burgers |
26 ≤ ? ≤ 28 | 7 |
28 ≤ ? ≤ 30 | 22 |
30 ≤ ? ≤ 32 | 36 |
32 ≤ ? ≤ 34 | 45 |
34 ≤ ? ≤ 36 | 33 |
36 ≤ ? ≤ 38 | 28 |
38 ≤ ? ≤ 40 | 4 |
Can it be assumed that the fat content of this grade of the burger is normally distributed?
Ho: distribution is normal probability distribution
Ha: distribution is not the normal probability
distribution
mean,µ= 33
std dev,s= 2.91
total observation , n = 175
lower bound | upper bound | percentage | observed frequency,O | expected frequency ( E )=5 | (O-E)²/E |
26.00 | 28.00 | 0.0429 | 7 | 7.50 | 0.034 |
28.00 | 30.00 | 0.1084 | 22 | 18.97 | 0.483 |
30.00 | 32.00 | 0.2143 | 36 | 37.50 | 0.060 |
32.00 | 34.00 | 0.2689 | 45 | 47.05 | 0.090 |
34.00 | 36.00 | 0.2143 | 33 | 37.50 | 0.539 |
36.00 | 38.00 | 0.1084 | 28 | 18.97 | 4.297 |
38.00 | 40.00 | 0.04 | 4 | 7.50 | 1.636 |
chi square test statistic,X² = Σ(O-E)²/E =
7.139
level of significance, α= 0.05
Degree of freedom=k-3 = 7
- 3 = 4
Critical value = 9.4877 [ Excel function:
=chisq.inv.rt(α,df) ]
decision: Fail to reject Ho
it can be assumed that the fat content of this grade of the burger is normally distributed