Question

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

Answer 1

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:

`z=(x-\mu)/((\sigma)/(√(n)))`

On substitution:

`\begin{gathered} z=(25-23.8)/((6.4)/(√(65))) \\ \\ z=1.512 \\ z\approx1.51(2\text{ }decimal\text{ }places) \end{gathered}`

So, the value of the test statistic = 1.51