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You measure the weights of 47 watermelons and find they have a mean weight of 77 ounces. Assume the population standard deviation is 14.4 ounces Based on this , construct a 95% confidence interval for the true population mean watermelon weight.___<μ<___ , use 3 decimal places.State in words what this means in context of the problem:

User Yinan
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Final answer:

The 95% confidence interval for the true population mean watermelon weight based on a sample mean of 77 ounces is 72.882 to 81.118 ounces. This interval expresses the range within which we are 95% sure that the true mean weight of the population of watermelons lies.

Step-by-step explanation:

To construct a 95% confidence interval for the true population mean watermelon weight, when the sample mean weight is 77 ounces, the population standard deviation is 14.4 ounces, and the sample size (n) is 47, we use the formula for the confidence interval of a population mean:

CI = μ ± Z*(σ/√n)

From the Z-table, a 95% confidence level corresponds to a Z-score (Z*) of approximately 1.96. Substituting the values, we get:

77 ± 1.96*(14.4/√47) = 77 ± 1.96*(2.101) = 77 ± 4.118

Calculating the range, we have:

Lower limit (LL) = 77 - 4.118 = 72.882

Upper limit (UL) = 77 + 4.118 = 81.118

Therefore, the 95% confidence interval is 72.882 < μ < 81.118.

In words, this means we are 95% confident that the true population mean weight of watermelons falls between 72.882 and 81.118 ounces.

User Daniel Hardt
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