Question

A recent survey indicated that the average amount spent for breakfast by business managers was $9.33 with a standard deviation of $0.24. It was felt that breakfasts on the East Coast were higher than $9.33. A sample of 81 business managers on the East Coast had an average breakfast cost of $9.41. At α=0.05, what is the test value?

Answer 1

Answer:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

Replacing the info given we got:

$t=\frac{9.41-9.33}{\frac{0.24}{\sqrt{81}}}=3$    

Step-by-step explanation:

Information given

$\bar X=9.41$ represent the sample mean

$s=0.24$ represent the sample standard deviation

$n=81$ sample size  

$\mu_o =9.33$ represent the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is higher than 9.33, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 9.33$  

Alternative hypothesis:$\mu > 9.33$  

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

Replacing the info given we got:

$t=\frac{9.41-9.33}{\frac{0.24}{\sqrt{81}}}=3$