Question

You think that the results below provided by a survey in regard to the average franchise investment information are incorrect:

• 43% for less than $100,000
• 41% for $100,000 and up
• 16% for people did provide any values

So, you randomly select 30 franchises and determine the necessary investment for each. The sample mean investment is $135,000 with a standard deviation of $30,000. Is there enough evidence to support your claim at alpha=0.05 that the average investment 143,260? (Use the p-value)

Answer 1

Givem Mean = $135000

standard deviation S = 30000

Null Hypothesis : $143260

Alternate Hypothesis : $143260

We will calculate the z-score

Z-score = ( - ) / S

= (143260 - 135000) / 30000

= 8260 / 30000

= 0.27533

The p-value for z-score of 0.27533 at a significance level of 0.05 for a two tailed test is 0.783086

Here we have used two-tailed test since we are not looking for the direction of the Mean, but just looking whether it is equal to or not equal to a particular value

If the p-value is less than value or Significance level, then we can reject the null hypothesis

Here p-value of 0.783086 is more than the value or Significance level of 0.05, hence we fail to reject the null hypothesis

So There is not enough sufficient evidence to support the claim that the average investment is $143,260 at alpha = 0.05