Question #347935

An association of City Mayors conducted a study to determine the average number of times a family went to buy necessities in a week. They found that the mean is 4 times in a week. A random sample of 20 families were asked and found a mean of 5 times in a week and a standard deviation of 2. Use 5% significance level to test that the population mean is not equal to 5. Assume that the population is normally distributed. What should be the decision for the hypothesis?

1
Expert's answer
2022-06-06T11:14:53-0400

The following null and alternative hypotheses need to be tested:

H0:μ=5H_0:\mu=5

H1:μ5H_1:\mu\not=5

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

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\}.

The z-statistic is computed as follows:


z=xˉμσ/n=552/20=0z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{5-5}{2/\sqrt{20}}=0

Since it is observed that z=0<1.6449=zc,|z|=0<1.6449=z_c, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is p=2P(z<0)=1,p=2P(z<0)= 1, and since p=1>0.05=α,p= 1>0.05=\alpha, it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean μ\mu

is different than 5, at the α=0.05\alpha = 0.05 significance level.


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