Please answer with details The population mean of weekly veg

Please answer with details

The population mean of weekly vegetable package sales is mu =3,000. The sample mean x-bar= s = 4,500, n=150. (1) Test H(O): mu =3,000, H(1): mu > 3,000 for x-bar =3,900, alpha =0.05: (2) Test H(0): mu =3,000, H(1): mu

Solution

(1) It is a right-tailed test.

Given a=0.05, the critical value is Z(0.05) =1.645 (from standard normal table)

The rejection region is if Z>1.645, we reject the null hypothesis.

The test statisitc is

Z=(xbar-mu)/(s/vn)

=(3900-3000)/(4500/sqrt(150))

=2.45

Since Z=2.45 is larger than 1.645, we reject the null hypothesis.

So we can conclude that mu>3000

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(2)It is a left-tailed test.

Given a=0.1, the critical value is Z(0.1) =-1.28 (from standard normal table)

The rejection region is if Z<-1.28, we reject the null hypothesis.

The test statisitc is

Z=(xbar-mu)/(s/vn)

=(2300-3000)/(4500/sqrt(150))

=-1.91

Since Z=-1.91 is less than -1.28, we reject the null hypothesis.

So we can conclude that mu<3000

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(3) It is a two-tailed test.

Given a=0.01, the critical value is Z(0.005) =2.58 or -2.58 (from standard normal table)

The rejection region is if Z>2.58 or Z<-2.58, we reject the null hypothesis.

The test statisitc is

Z=(xbar-mu)/(s/vn)

=(4000-3000)/(4500/sqrt(150))

=2.72

Since Z=2.72 is larger than 2.58, we reject the null hypothesis.

So we can conclude that mu not equal to 3000