Consider the following hypothesis test: h0: u u is less than or equal to 50ha: u > 50a sample of 60 is used and the population standard deviation is 8. use the critical value approach to state your conclusion for each of the following sample results. use = .05.a. with ? ? x = 52.5, what is the value of the test statistic (to 2 decimals)? can it be concluded that the population mean is greater than 50? b. with ? x? = 51, what is the value of the test statistic (to 2 decimals)? can it be concluded that the population mean is greater than 50? c. with ? x? = 51.8, what is the value of the test statistic (to 2 decimals)? can it be concluded that the population mean is greater than 50?

Step-by-step explanation:

Because we know the population standard deviation, we must use a Z-test.

Null hypothesis (H0): μ=50  

Alternative hypothesis (H1): μ> 50

The decision rule is:

z-statistic< Z(z-student table (alpha/2)) --> You must accept the null hypothesis

z-statistic > Z (z-student table(alpha/2)) --> You must reject the null hypothesis

z-statistic formula:

z= (xbar-m)/(σ/(sqrt(n)))

xbar: sample mean

m: hypothesized value

σ: population standard deviation

n: number of observations

a)

z=(52.5-50)/(8/sqrt(60))

z= 2.42  

The z-distribution table statistic at 2,5% (alpha/2) significance level is: 1.96 Because z-statistic is greater than the z-table value, we must reject the null hypothesis. It can be concluded that the population mean is greater than 50.

b)

z=(51-50)/(8/sqrt(60))

z= 0.96

The z-distribution table statistic at 2,5% (alpha/2)significance level is the same: 1.96 Because z-statistic is less than the z-table value, it cannot be concluded that the population mean is greater than 50.

c)

z=(51.8-50)/(8/sqrt(60))

z= 1.74

The z-distribution table statistic at 2,5%(alpha/2) significance level is the same: 1.96 Because z-statistic is less than the z-table value, it cannot be concluded that the population mean is greater than 50.

probability for j is 1/4

while

a is 1/6