Question

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.

From a random sample of 74 ​dates, the mean record high daily temperature in a certain city has a mean of 85.22 degrees F. Assume the population standard deviation is 13.97 degrees F.

Answer 1

Answer:

Very confused! at 95% confidence level, the true population mean will be between 134.33 and 145.67.

Step-by-step explanation:

A random sample of 45 home theater systems has a mean price of ​$140.00

Assume the population standard deviation is ​$19.40

population standard deviation is 19.40

sample size is 45.

standard error of the sample mean is 19.40 / sqrt(45) = 2.89198

at 90% confidence level, critical z is plus or minus 1.645.

at 95% confidence level, critical z is plus or minus 1.96.

critical raw score will be sample mean plus or minus critical z-score * standard error of the mean.

your sample mean is 140.

your standard error of the mean is 2.89198

at 90% confidence level, your critical raw score will be 140 plus or minus 1.645 * 2.89198.

that becomes 140 plus or minus 4.76 rounded to the nearest penny.

at 90% confidence level, the true population mean will be between 135.24 and 144.76.

at 95% confidence level, your critical raw score will be 140 plus or minus 1.96 * 2.89198.

that becomes 140 plus or minus 5.67 rounded to the nearest penny.

Answer 2

Using the z-distribution, it is found that:

• The 90% confidence interval for the population mean is (82.55, 87.89). It means that we are 90% sure that the true population mean is between these two values.

• The 95% confidence interval for the population mean is (82.04, 88.40). It means that we are 95% sure that the true population mean is between these two values.

• The higher the confidence level, the higher the critical value is, and thus, a higher confidence level leads to a wider interval.

We are given the standard deviation for the population, and thus, the z-distribution is used.

The information is:

• Sample mean of $\overline{x} = 85.22$.
• Population standard deviation of $\sigma = 13.97$.
• Sample size of 74, thus $n = 74$.

A z-confidence interval is given by:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

We have to find the critical value, which is z with a p-value of $\frac{1 + \alpha}{2}$, in which $\alpha$ is the confidence level.

For the 90% confidence interval, $\alpha = 0.9$, thus, z with a p-value of $\frac{1 + 0.9}{2} = 0.95$, which means that it is z = 1.645.

Then, the 90% confidence interval is:

$\overline{x} - z\frac{\sigma}{\sqrt{n}} = 85.22 - 1.645\frac{13.97}{\sqrt{74}} = 82.55$

$\overline{x} + z\frac{\sigma}{\sqrt{n}} = 85.22 + 1.645\frac{13.97}{\sqrt{74}} = 87.89$

The 90% confidence interval for the population mean is (82.55, 87.89). It means that we are 90% sure that the true population mean is between these two values.

For the 95% confidence interval, $\alpha = 0.95$, thus, z with a p-value of $\frac{1 + 0.95}{2} = 0.975$, which means that it is z = 1.96.

Then, the 95% confidence interval is:

$\overline{x} - z\frac{\sigma}{\sqrt{n}} = 85.22 - 1.96\frac{13.97}{\sqrt{74}} = 82.04$

$\overline{x} + z\frac{\sigma}{\sqrt{n}} = 85.22 + 1.96\frac{13.97}{\sqrt{74}} = 88.40$

The 95% confidence interval for the population mean is (82.04, 88.40). It means that we are 95% sure that the true population mean is between these two values.

The higher the confidence level, the higher the critical value is, and thus, a higher confidence level leads to a wider interval.

A similar problem is given at https://brainly.com/question/22596713