The attention span of little kids (ages 3–5) is claimed to be normally distributed with a mean of 15 minutes and a standard deviation of 4 minutes. a test is to be performed to decide if the average attention span of these kids is really this short or if it is longer. you decide to test the hypotheses h0: μ = 15 versus ha: μ > 15 at the 5% significance level. a sample of 10 children will watch a tv show they have never seen before, and the time until they walk away from the show will be recorded. if, in fact, the true mean attention span of these kids is 18 minutes, what is the probability of a type ii error?

P(Type ll error) = 0.2327

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 minutes

Sample size, n = 10

Alpha, α = 0.05

Population standard deviation, σ = 4 minutes

First, we design the null and the alternate hypothesis

We use One-tailed z test to perform this hypothesis.

Formula:

Putting the values, we get,

Type ll error is the error of accepting the null hypothesis when it is not true.

P(Type ll error)

Calculating value from the z-table we have,

Thus,

P(Type ll error) = 0.2327

answer: a

step-by-step explanation:

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.