Kat is interested in getting chickens so she can have fresh eggs. Before she buys her chickens, she wants to find the mean number of eggs each one will lay. She begins by randomly selecting 40 chickens from a large poultry farm and counts how many eggs each one lays within one month. She finds the mean to be 24.8 eggs with a standard deviation of 6.9 eggs. Which of the following is the 90% confidence interval for the true mean number of eggs chickens from this poultry farm lay in one month?

Find the t-table here.

(21.85, 27.75)
(22.59, 27.01)
(22.96, 26.64)
(23.38, 26.22)

The confidence interval of eggs chickens from this poultry farm lay in one month will be (22.96, 26.64). Then the correct option is C.

What is the confidence interval of the sample?

If the sample size is given to be n < 30, then for finding the confidence interval for mean of population from this small sample, we use t-statistic.

Let the sample mean given as μ and

The sample standard deviation σ, and

The sample size = n, and

Then the heights would be outside the 90% confidence interval will be

For 90% confidence interval, we have z = 1.645

⇒ μ ± z (σ/√n)

Kat is interested in getting chickens, so she can have fresh eggs.

Before she buys her chickens, she wants to find the mean number of eggs each one will lay.

She begins by randomly selecting 40 chickens from a large poultry farm and counts how many eggs each one lays within one month.

She finds the mean to be 24.8 eggs with a standard deviation of 6.9 eggs.

The 90% confidence interval for the true mean number of eggs chickens from this poultry farm lay in one month will be

⇒ μ ± z (σ/√n)

⇒ 24.8 ± 1.645 (6.9/√40)

⇒ 24.8 ± 1.84

Then the confidence interval of eggs chickens from this poultry farm lay in one month will be (22.96, 26.64).

Then the correct option is C.

Learn more about confidence interval of sample here:

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