One hundred random samples, each of size 25, are obtained from the normal distribution with mean 0 and standard deviation 1 using minitab. subsequently, the 1-sample z procedure in minitab is used (with the same confidence level) to obtain a confidence interval from each sample. out of the 100 intervals thus obtained, 89 include the number 0.
estimate the confidence level (in percentage terms) used to generate the 100 intervals using a 95% confidence interval.

- 0.196 < μ < +0.196

Step-by-step explanation:

n = 100

σ = 1

X bar (x⁻) = 0

1 - ∝ =95%

μ = ?

= X bar (x⁻) ± Z ∝/₂ * σ/√n

Finding ∝:

To find ∝, we will convert the confidence interval i.e. 95% into decimal by dividing it with 100 and subtracting it from 1.

1 - ∝ = 95%

1 - ∝ = 95/100

1 - ∝ = 0.95

∝ = 1 - 0.95

∝ = 0.05 (1)

We have to calculate ∝/2 so dividing both sides of (1) with 2

∝/2 = 0.05/2

∝/2 = 0.025

We will find the value of 0.025 in z-table. The value is:

Z ∝/2 = 1.96

For further calculations, we will use the value of Z∝/2.

Putting all values in the formula.

= 0 ± 1.96 * 1 / √100

We can also write this in the following form:

0 - 1.96/ √100 < μ < 0 + 1.96 / √100

- 1.96 / 10 < μ < 1.96 / 10   ∴ √100 = 10

- 0.196 <  μ < + 0.196

i think the answer is seven, for if i'm being useless °ω°

step-by-step explanation: