Question

A marketing consultant is hired by a major restaurant chain wishing to investigate the preferences and spending patterns of lunch customers. The CEO of the chain hypothesized that the average customer spends at least $13.50 on lunch. A survey of 25 customers sampled at one of the restaurants found the average lunch bill per customer to be x¯= $14.50. Based on previous surveys, the restaurant informs the marketing manager that the standard deviation is ???? = $3.50. To address the CEO’s conjecture, the marketing manager decides to carry out a test of hypothesis. The null hypothesis is given by H0 : ???? = 13.50. The hypothesis test is set up to provide the marketing manager with:
a. proof that the null hypothesis is true.
b. evidence against the null hypothesis in favor of the alternative.
c. evidence against the alternative hypothesis in favor of the null hypothesis.
d. evidence in favor of the null hypothesis.

Answer 1

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

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

substitute all values , we get

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