Question

A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds. Round all answers to the nearest hundredth.Conclusion: We estimate with 90% confidence that the mean weight of all heads of lettuce in pounds is between?

Answer 1

To construct a 95% confidence interval for the mean weight of the heads of lettuce, we use the formula: Confidence interval = sample mean ± (critical value) × (standard deviation / square root of sample size). Given the sample mean weight, standard deviation, and sample size, we can calculate the confidence interval and estimate the mean weight with 95% confidence.

Step-by-step explanation:

To construct a 95% confidence interval for the population mean weight of the heads of lettuce, we can use the formula:

Confidence interval = sample mean ± (critical value) × (standard deviation / square root of sample size)

Given that the sample mean weight is 2.2 lb, the standard deviation is 0.1 lb, and the sample size is 20, we can calculate the confidence interval as follows:

Confidence interval = 2.2 ± (1.96) × (0.1 / sqrt(20))

Rounding to the nearest hundredth, the confidence interval is approximately [2.16, 2.24] lb.

This means that we estimate with 95% confidence that the mean weight of all heads of lettuce in pounds is between 2.16 lb and 2.24 lb.

Answer 2

Given:

- A sample of 20 heads of lettuce was selected.

- The population distribution of head weight is normal.

- The mean weight was:

`\mu=2.2lb`

with a Standard Deviation of 0.1 pounds.

- The population standard deviation is known to be 0.2 pounds.

By definition, the z-statistics for a 90% Confidence Interval is:

`z_{(a)/(2)}=1.645`

The formula for calculating the Confidence Interval for the Mean is:

`x\pm z_{(a)/(2)}\frac{\sigma}{\sqrt[]{n}}`

Where "x" is the sample mean, "n" is the sample size, the Standard Deviation is σ, and the z-statistics, in this case, is the one shown before.

In this case:

`\begin{gathered} n=20 \\ x=2.2 \\ \sigma=0.2 \end{gathered}`

Therefore, substituting values into the formula and evaluating, you get these two values:

`2.2\pm(1.645)(_{}\frac{0.2}{\sqrt[]{20}})`
`\Rightarrow\begin{cases}2.2+(1.645)(_{}\frac{0.2}{\sqrt[]{20}})\approx2.27 \\ \\ 2.2-(1.645)(_{}\frac{0.2}{\sqrt[]{20}})\approx2.13\end{cases}`

Hence, the answer is:

`2.13\text{ and }2.27`