Question
In: Statistics and Probability
A farmer weighs 10 randomly chosen watermelons from his farm and he obtains the following values...
A farmer weighs 10 randomly chosen watermelons from his farm and
he obtains the following values (in lbs):
7.72,9.58,12.38,7.77,11.27,8.80,11.10,7.80,10.17,6.00
Assuming that the weight is normally distributed with mean μ and
and variance σ2, find a 95% confidence interval for μ.
Solutions
Expert Solution
[first we calculating mean and variance to contruct the 95% confidence interval]
Mean = (7.72 + 9.58 + 12.38 + 7.77 + 11.27 + 8.8 + 11.1 + 7.8 +
10.17 + 6) / 10
Mean = 92.59/ 10
Mean = 9.26
Calculating Population Variance
We can calculate the population variance using the below
formula:
Put the values in the formula,
= 1.89
Calculating 95% confidence interval for mean:
Mean, x bar = 9.26,
Standard Deviation, = 1.89,
Sample Size, n = 10,
z critical value of 95% confidence interval = 1.96
Now, using the formula
Put the values in the above formula:
= 8.0886, 10.4314
So, Confidence Interval = [8.0886, 10.4314]
orchestra answered 2 years agoRelated Solutions
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