Question

A random sample of 78 eighth grade​ students' scores on a national mathematics assessment test has a mean score of 284.
This test result prompts a state school administrator to declare that the mean score for the​ state's eighth graders on this exam is more than 275.

Assume that the population standard deviation is 32. At α=0.04​, is there enough evidence to support the​ administrator's claim.a)
H0​:μ≤275Ha​:μ>275​(claim)

b) find the standarized test statistic z and its corresponding areac) find the p valued) describe whether to reject or fail the null hypothesise) interpret your decision in the context of the original claim

Answer 1

Answer:

a) In the step-by-step-explanation

b) z(s) = -2.46     corresponding area is 0.0069      

P [  μ₀   > 275   ]  is 0.0069      or     0.69 %

We reject  H₀

Step-by-step explanation:

Normal distribution

Random sample

size sample 78   =  n

population standard deviation σ = 32

The school administrator declare that mean score is more (bigger than)

275. So the hypothesis test should be:

H₀  null hypothesis                               μ₀   > 275           and

Hₐ  alternative hypothesis                   μ₀   < 275

Is one tail test with  α  = 0,04 from tables we have  z(c) = - 176

We proceed to compute z(s)

z(s) = [ (μ -  μ₀) /( σ /√n) ]       ⇒  z(s) = (-  9 *√78 )/ 32

z(s) = - (9*8.83)/32  

z(s) = - 2.46    corresponding area   is  0,0069

P [ z > 275 ] = 0.0069          or     0.69 %

The value for z(s)  = - 2.46  is smaller than the critical value mentioned in problem statement z(c) = - 1.74  , the z(s) is in the rejection zone

Therefore we reject H₀