contestada

A recent survey reported that small businesses spend 25 hours a week marketing their business.A local Chamber of Commerce claims that small businesses in the area are not growing because these businesses are spending less than 25 hours a week on marketing. The chamber conducts a survey of 65 small businesses within their state and finds that the average amount of time spent on marketing is 23.8 hours a week.assuming that the population standard deviation is 6.4 hours, is there a sufficient evidence to support the chamber of commerce claim at the 0.05 level of significance?Compute the value of the test statistic. Round answer to two decimal places

A recent survey reported that small businesses spend 25 hours a week marketing their businessA local Chamber of Commerce claims that small businesses in the are class=

Respuesta :

Given;

supposed number of hours, x = 25hours

Number of business, n = 65

average time spent, μ= 23.8hours

standard deviation, σ = 6.4

given propability,α = 0.05

Let us first find the value of z;

Recall that:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

On substitution:

[tex]\begin{gathered} z=\frac{25-23.8}{\frac{6.4}{\sqrt{65}}} \\ \\ z=1.512 \\ z\approx1.51(2\text{ }decimal\text{ }places) \end{gathered}[/tex]

So, the value of the test statistic = 1.51