a random sample of 130 students is chosen from a population of 4,500 students. if the mean iq in the sample is 120 with a standard deviation of 5, what is the 99% confidence interval for the students' mean iq score? answers given below: 118.87−121.13 107.12−132.88 125−135 115−125

Respuesta :

[tex]C.I.=120 \pm z_{ \alpha /2} \frac{ \sigma }{ \sqrt{n} } \\ =120 \pm 2.58 \times \frac{5}{ \sqrt{130} } \\ =120 \pm 1.13 \\ =118.87 - 121.13[/tex]

Answer with explanation:

Total Sample Size = 4,500 Students

Sample Chosen(n) = 130 students

Mean IQ of the Sample,[tex]\bar{X} = 120[/tex]

Standard Deviation, [tex]\sigma=5[/tex]

To Calculate 99% confidence interval for the students' mean IQ score, we will calculate,

[tex]Z_{99 percent}=2.58[/tex]

Formula for Confidence Interval

[tex]=\bar{X} \pm Z_{99 percent}\times \frac{\sigma}{\sqrt{n}}\\\\=120 \pm 2.58 \times \frac{5}{\sqrt{130}}\\\\=120 \pm \frac{12.9}{11.401}\\\\ =120 \pm 1.1314\\\\=120 \pm 1.13[/tex]

The Value of Confidence interval will lie between

⇒  120 - 1.13 to 120 +1.13

⇒118.87 to 121.13

Option A:  118.87−121.13