The surface hardness measurement of a floor has been monitor

The surface hardness measurement of a floor has been monitored for a long period of time. It has been determined that the hardness measurement has a normal distribution with mean 4.5 and standard deviation 1.5. A new production team that makes the floors needs to be monitored. Let x-bar = 3.9 denote the sample mean of hardness measurements from 25 floor samples.

Let =0.10.

Test the hypotheses H₀: µ = 4.5 vs. H_a: µ 4.5 using the

1. Rejection Region Approach and the Confidence Interval Approach

Solution

here H0: mu=4.5   and Ha: mu is not equal to 4.5

rejection region approach

here the test statistic is T=(xbar-mu)*sqrt(n)/sigma which under H0 follows a standard normal distribution

where xbar=sample mean=3.9 mu=4.5 n=sample size=25 sigma=population standard deviation=1.5

alpha is given as 0.10 so alpha/2=0.05 so critical value is 1.96 where critical value is the upper alpha/2 point of a standard normal distribution.

we reject H0 iff | observed value of T|>critical value

now observed value of T=(3.9-4.5)*sqrt(25)/1.5=-2

so |-2|=2>1.96

hence based on the given data at hand we reject the null hypothesis and conclude that the hardness measurement does not have mean 4.5

confidence interval approach

the 90% confidence interval is

[xbar-sigma/sqrt(n)*critical value(at alpha/2),xbar+sigma/sqrt(n)*critical value(at alpha/2)]

=[3.9-1.5*1.96/5,3.9+1.5*1.96/5]=[3.312,4.488]

now the interval does not contain 4.5

hence hence based on the given data at hand we reject the null hypothesis and conclude that the hardness measurement does not have mean 4.5