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You think (claim) that the average franchise investment information shown in
the graph above is incorrect, 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 a = 0.05?

You think claim that the average franchise investment information shown in the graph above is incorrect so you randomly select 30 franchises and determine the n class=

Respuesta :

Ok I’m not trying anything for real but it’s for a couple days I’m going

Using t-test, there isn't an enough evidence to support our claim that investment average is $135,000 at α = 0.05.

What is t - test?

The one sample t test compares the mean of your sample data to a known value (population mean in our question, i.e. , we wish to compare the sample mean with the population mean). One sample t test should be used when you the population standard deviation is not known or the sample size is very small.

Assumptions of the t-test for the test to be valid:-

  • Data is independent.
  • Data is collected randomly. For example, with simple random sampling.
  • The data is approximately normally distributed.

Null hypothesis:  μ = $143,260

Alternative hypothesis: μ < $143,260

Calculating test statistic.

n = 30

sample mean = $135,000

sample standard deviation = $30,000

[tex]t_{0} = \frac{sample\ mean - population\ mean}{\frac{standard\ deviation}{\sqrt{sample\ size} } }[/tex]

[tex]t_{0} = \frac{135000 - 143260}{\frac{30000}{\sqrt{30} } }[/tex]

|to| = 1.508

Test criteria:

level of significance = 0.05

t(0.05, 29) = 1.6991

Since, |to| = 1.508 < t(0.05, 29) = 1.6991, null hypothesis cannot be rejected.

Learn more about t-test here

https://brainly.com/question/15870238

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