106k views
0 votes
You measure 44 watermelons' weights, and find they have a mean weight of 50 ounces. Assuméthe

c»putation standard deviation is 10.5 ounces. Based on this, construct a 99% confidence interval for the
true population mean watermelon weight.

1 Answer

2 votes

Final answer:

To construct a 99% confidence interval for the true population mean watermelon weight, you can use the sample mean and standard deviation. Using the formula, calculate the critical value based on the desired confidence level and sample size. Then, calculate the confidence interval using the critical value, sample mean, and standard deviation.

Step-by-step explanation:

To construct a 99% confidence interval for the true population mean watermelon weight, we can use the formula:

Confidence interval = sample mean ± (critical value) × (standard deviation of sample mean)

Given that we have a sample mean weight of 50 ounces and a standard deviation of 10.5 ounces, we need to find the critical value for a 99% confidence level. Since our sample size is 44, we can use a t-distribution instead of a z-distribution.

Using a t-table or calculator, we find that the critical value for a t-distribution with 43 degrees of freedom at a 99% confidence level is approximately 2.6927.

Now we can calculate the confidence interval:

Confidence interval = 50 ± 2.6927 × (10.5 / √44)

Confidence interval = (43.04, 56.96)

Therefore, we can be 99% confident that the true population mean watermelon weight is between 43.04 ounces and 56.96 ounces.

User Tomty
by
7.3k points