Question

A recent study claims that business travelers spend an average of $39 per day for meals. A sample of 15 business travelers found that they had spent an average of $42 per day with a standard deviation of $3.78. If α=0.05, what is the test value?

Answer 1

Answer:

The test statistic value 't' = 3.074

Step-by-step explanation:

Step(i):-

Given sample size 'n' = 15

mean of the Population 'μ' =  $39

mean of the sample x⁻ =  $42

standard deviation of the sample 's' =  $3.78

Degrees of freedom ν = n-1 = 15-1 =14

$t_{\frac{\alpha }{2} } = 1.769$

Null hypothesis : There is no significance difference between the means

H₀ :  x⁻  =  'μ'

Alternative hypothesis : There is significance difference between the means

H₀ :  x⁻ ≠  'μ'

Test statistic  

$t = \frac{x^{-} -mean}{\frac{s}{\sqrt{n} } }$

$t = \frac{42 -39}{\frac{3.78}{\sqrt{15} } }$

t = 3.074

The test value of t-statistic  t = 3.074

The calculated value  t = 3.074 > 1.769 at 0.05 level of significance

null hypothesis is rejected

Alternative hypothesis is accepted

Final answer:-

There is significance difference between the means