Answer:
There is no sufficient evidence to conclude that population mean number of times that adults go out for dinner each week is less than 1.5
Step-by-step explanation:
From the question we are told that
The population mean is less than 1.5
The sample size is n = 7
The sample data is 2, 0, 1, 5, 0, 2, and 3
Generally the sample mean is mathematically represented as
[tex]\= x = \frac{ \sum x_i }{ n }[/tex]
=> [tex]\= x = \frac{ 2 + 0 + 1 + 5 + 0 + 2 + 3 }{ 7 }[/tex]
=> [tex]\= x = 1.857 [/tex]
Generally the standard deviation is mathematically represented as
[tex]\sigma = \sqrt{\frac{\sum (x_i - \= x)^2}{n} }[/tex]
=> [tex]\sigma = \sqrt{\frac{ (2 - 1.857)^2 + (0 - 1.857)^2 +\cdots (3 - 1.857)^2}{7} }[/tex]
=> [tex]\sigma = 1.773[/tex]
The null hypothesis is [tex]H_o : \mu = 1.5[/tex]
The alternative hypothesis is [tex]H_a : \mu < 1.5[/tex]
Generally the test statistics is mathematically represented as
[tex]z = \frac{\= x - \mu }{ \frac{\sigma }{\sqrt{n} } }[/tex]
=> [tex]z = \frac{\= x - 1.857 }{ \frac{1.773 }{\sqrt{7} } }[/tex]
=>[tex]z = 0.53[/tex]
Generally the p-value is mathematically represented as
[tex]p-value = P(z < 0.53 )[/tex]
From the z-table
[tex]P(z < 0.53 ) = 0.70194[/tex]
So
[tex]p-value = 0.70194[/tex]
From the values obtained we see that [tex]p-value > 0.05[/tex]
Here there is no sufficient evidence to conclude that population mean number of times that adults go out for dinner each week is less than 1.5
