we have the converted data is
| converted |
| 5.833333 |
| 5.416667 |
| 5.25 |
| 5.583333 |
| 5.833333 |
| 5.166667 |
| 6 |
| 5.666667 |
| 5.333333 |
| 5.833333 |
| 5.333333 |
| 5.583333 |
| 5.2 |
| 5.833333 |
| 5.291667 |
| 5.566667 |
| 5.266667 |
| 5.266667 |
| 4.916667 |
| 5.916667 |
| 5.083333 |
| 5.083333 |
| 5.166667 |
| 4.916667 |
| 5.25 |
| 5.333333 |
Ans a ) the histogram for the converted data is

Ans 2 ) using minitab>stat>basic stat>display descriptive stat
we have
Descriptive Statistics: converted
Variable Mean StDev Median
converted 5.4202 0.3134 5.3333
mean = 5.4202
Standard deviaation = 0.3134
median = 5.3333
Ans 3 ) the null and alternative hypothesis is
Ho : u = 5.5
Ha:u ≠ 5.5 (two tailed )
using minitab>stat>basic stat>one sample t
we have
One-Sample T: data
Test of μ = 5.5 vs ≠ 5.5
Variable N Mean StDev SE Mean 95% CI T P
data 10 5.700 1.194 0.377 (4.846, 6.554) 0.53 0.609
the value of test stat t = 0.53
p value = 0.609
since p value is greater than 0.05 so we do not reject Ho and conclude that average chicken weight is 5.5
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