Question #334273

 the mean monthly Internet bill is $32.79 per household. A sample of 50 households in the Belize District showed a sample mean of $30.63. Use a population standard deviation $5.60.

  

 What is the value of the test statistic? 

c) What is the p-value?

d) At α = .05, what is your conclusion?


1
Expert's answer
2022-04-28T08:41:26-0400

The following null and alternative hypotheses need to be tested:

H0:μ=32.79H_0:\mu=32.79

H1:μ32.79H_1:\mu\not=32.79

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

b) The z-statistic is computed as follows:


z=xˉμσ/nz=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}

c) The p-value is p=2P(Z<z),z0p=2P(Z<z), z\le0 or p=2P(Z>z),z0.p=2P(Z>z), z\ge0.


d) Based on the information provided, the significance level is α=0.05,\alpha = 0.05, and the critical value for a two-tailed test is zc=1.96.z_c = 1.96.

The rejection region for this two-tailed test is R={z:z>1.96}.R = \{z: |z| > 1.96\}.


z=30.6332.795.60/502.7274z=\dfrac{30.63-32.79}{5.60/\sqrt{50}}\approx-2.7274

Since it is observed that z=2.7274>1.96=zc,|z| = 2.7274 >1.96= z_c , it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p=2P(Z<2.7274)=0.006384,p=2P(Z<-2.7274)=0.006384, and since p=0.006481<0.05=α,p=0.006481<0.05=\alpha, it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ\mu is different than 32.79, at the α=0.05\alpha = 0.05 significance level.


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