Question

I really need help with my Confidence Intervals for the Mean homework, can someone please help?

1. You measure 34 dogs' weights, and find they have a mean weight of 52 ounces. Assume the population standard deviation is 10.2 ounces. Based on this, construct a 99% confidence interval for the true population mean dog weight.

Give your answers as decimals, to two places

________________ ± ______________ ounces

2. Assume that a sample is used to estimate a population mean μμ. Find the margin of error M.E. that corresponds to a sample of size 17 with a mean of 20.8 and a standard deviation of 20.4 at a confidence level of 80%.

Report ME accurate to one decimal place because the sample statistics are presented with this accuracy.
M.E. = _____________

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

3. The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systolic blood pressure is 20.8 for a sample of size 1050 and standard deviation 20.4.
Estimate how much the drug will lower a typical patient's systolic blood pressure (using a 99% confidence level).
Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

____________ < μ < ____________

Answer should be obtained without any preliminary rounding.

4. You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures:

23.9
25.5
13.5
23.2
23.5
23.1
16.7
-12.1
42.4

Find the 90% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).

90% C.I. = ______________

Answer should be obtained without any preliminary rounding.

Answer 1

1) At 99% confidence level, the critical value is z_0.005 = 2.58

The 99% confidence interval is

T士 20.005 * -

10.2 = 52 +2.58 + v34

= 52±4.51

2) DF = 17 - 1 = 16

At 80% confidence level, the critical value is t^* = 1.337

S M.E = ***

           

= 1.337*

           

3) At 99% confidence level, the critical value is z_0.005 = 2.58

The 99% confidence interval is

S T = 20.005 * - n

20.4 = 20.8 +2.58 + 1050

= 20.8 +1.6

= 19.2.22.4

19.2 <μ< 22.4

4) $\bar x$ = 19.967

    s = 14.417

DF = 9 - 1 = 8

At 90% confidence level, the critical value is t^* = 1.860

The 90% confidence interval is

.. * ¥于C

14.417 = 19.967 1.86 * = V9

= 19.96748.94

= 11.027, 28.907

= 11.0, 28.9