Question

Below is the output from a study to see whether there is evidence of a difference in the mean amounts of time required to reach a customer service representative between two hotels. Assume that the population variances in the amount of time for the two hotels are not equal. What is the smallest level of significance at which the null hypothesis will still not be rejected?

Hotel 1 Hotel 2
Mean 2.214 2.0115
Variance 2.951657 3.57855
Observations 20 20
Hypothesized Mean Difference 0
df 38
t Stat 0.354386
P(T<=t) one-tail 0.362504
t Critical one-tail 1.685953
P(T<=t) two-tail 0.725009
t Critical two-tail 2.024394


Required:
a. State the null and alternative hypotheses for testing if there is evidence of a difference in the variabilities of the amount of time required to reach a customer service representative between the two hotels.
b. What is the value of the test statistic for testing if there is evidence of a difference in the variabilities of the amount of time required to reach a customer service representative between the two hotels? Explain how you obtain your answer.
c. What is the critical value for testing if there is evidence of a difference in the variabilities of the amount of time required to reach a customer service representative between the two hotels at the 5% level of significance?

Answer 1

a).
`$\text{Set up hypothesis}$`

`$\text{Null hypothesis}$` : There is no significance between them.

`$H_0:\mu_1=\mu_2$`

`$\text{Alternate hypothesis}$` : There is significance between them.

`$H_1:\mu_1!=\mu_2$`

b). Test statics

`$X(\text{mean}) = 2.214$`

Standard deviation
`$(s.d1)=1.718; \text{Number}(n_1)=20$`

`$Y(\text{mean})=2.0115$`

Standard deviation
`$(s.d2)=1.8917; \text{Number}(n_2)=20$`

We use the test statics :

`$t=\frac{X-Y}{\sqrt{(s.d1^2)/(n_1)+(s.d2^2)/(n_2)}}$`

`$t_0=\frac{2.214-2.0115}{\sqrt{(2.95152)/(20)+(3.57853)/(20)}}$`

`$t_0=0.35$`

`$|t_0|=0.35$`

c. The critical value of |t| with minimum
`$(n_1-1)$` , i.e. 19 d.f is
`$2.093$`

We got
`$|t_0|$` = 0.35439 and |t| = 2.093