contestada

Assume a 2015 Gallup Poll asked a national random sample of 483 adult women to state their current weight. Assume the mean weight in the sample was x¯=159.

We will treat these data as an SRS from a normally distributed population with standard deviation ????=33 pounds.

Give a 99% confidence interval for the mean weight of adult women based on these data. Enter the upper and lower values of your confidence interval into the spaces provided rounded to two decimal places.

lower value =__________pounds

upper value =__________pounds

Respuesta :

Answer:

Lower value = 155.13 pounds

and,

upper value = 162.87 pounds

Step-by-step explanation:

Data provided in the question:

Sample size, n = 483 adults

Sample mean [tex]\bar{x}[/tex]= 159

Standard deviation, σ = 33 pounds

Confidence level = 99%

Now,

z value for 99% confidence level is 2.58

Therefore,

Margin of error, (E) = [tex]\frac{\sigma}{\sqrt{n}}\times z[/tex]

thus,

Margin of error, (E) = [tex]\frac{33}{\sqrt{483}}\times2.58[/tex]

or

Margin of error, (E) = 3.87

Therefore,

Lower value = Mean - Margin of error

= 159 - 3.87 = 155.13 pounds

and,

upper value = Mean + Margin of error

= 159 + 3.87 = 162.87 pounds

Otras preguntas