Respuesta :
Answer:
t = 1.399<1.710 ( from tabulated value)
we will accepted the null hypothesis
the marketing manager decides to carry out a test of hypothesis
H 0 : μ = 13.50
Step-by-step explanation:
Step1 :-
Given sample size 'n' = 25 customers
The CEO of the chain hypothesized that the average customer spends at least $13.50 on lunch.
therefore the population mean 'μ' = $13.50
A survey of 25 customers sampled at one of the restaurants found the average lunch bill per customer to be ¯ x = $ 14.50
sample mean ¯ x = $ 14.50
the restaurant informs the marketing manager that the standard deviation is σ = $ 3.50
Population standard deviation σ = $ 3.50
Step 2:-
we will use' t' test because of given small sample size is n=25
[tex]t = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n-1} } }[/tex]
Null hypothesis: H₀ : μ= $ 13.50
Alternative hypothesis :H₁:μ ≠ $ 13.50
Degrees of freedom γ = n-1 = 25-1 =24
The test statistic 't' is
[tex]t = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n-1} } }[/tex]
substitute all values , we get
[tex]t = \frac{14.50 -13.50}{\frac{3.50}{\sqrt{25-1} } }[/tex]
[tex]t = \frac{1}{0.7144}[/tex]
t = 1.399
This is calculated value t = 1.399
The tabulated value of t are 5% level with 24 degrees of freedom
t = 1.710
Step 3:-
t = 1.399<1.710
Since calculated value < The tabulated value of t are 5% level with 24 degrees of freedom.
we will accepted the null hypothesis
the marketing manager decides to carry out a test of hypothesis
