Question

2. You read a study stating the average franchise investment is $143,260, st ? $24,000. You think this information is incorrect, so you randomly select 30 franchises and determine the sample mean investment is $135,000, standard deviation of $25,000. Is there enough evidence to support your claim at ?-0.05? Use the P-value method to test the hypothesis.

Answer 1

Hypotheses are:

$H_{0}:\mu=143260 \text{ (null hypothesis)}$

$H_{a}:\mu\neq 143260 \text{ (alternative hypothesis)}$

Test is two tailed.

Given information:

$\sigma=24000,n=30,\bar{x}=135000$

Since population standard deviation is known so single sample z test will be used.

The test statistics is

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{135000-143260}{24000/\sqrt{30}}\approx -1.89$

Using excel function "=2NORMSDIST(-1.89)" the p-value of the test is

p-value = 0.0588

Since p-value is greater than 0.05 so we fail to reject the null hypothesis.