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 1 year ago

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