contestada

A company specializing in building robots that can clean your house has found that the average amount of time kids spend cleaning their houses is about 2 hours per week. If their sample size was 1000 randomly chosen kids and the standard deviation was 0.3 hours, what is the margin of error for a confidence interval of 95%?

Respuesta :

MathPhys

Answer:

0.588

Step-by-step explanation:

Margin of error is:

ME = CV × SE

where CV is the critical value (z score or t score), and SE is the standard error (standard deviation of the sample).

Since n is large, we can approximate as a normal distribution.  At 95% confidence level, z = 1.96.

Therefore, the margin of error is:

ME = 1.96 × 0.3

ME = 0.588

Using the z-distribution, it is found that the margin of error for a 95% confidence interval was of 0.0186.

What is a z-distribution confidence interval?

The interval is:

[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]

In which:

  • [tex]\overline{x}[/tex] is the sample mean.
  • z is the critical value.
  • n is the sample size.
  • [tex]\sigma[/tex] is the standard deviation for the population.

The margin of error is given by:

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.

The standard deviation for the population and the sample size are given, respectively, by:

[tex]\sigma = 0.3, n = 1000[/tex]

Hence, the margin of error is given by:

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]M = 1.96\frac{0.3}{\sqrt{1000}}[/tex]

M = 0.0186.

More can be learned about the z-distribution at https://brainly.com/question/25890103

#SPJ2

Otras preguntas