Respuesta :
Answer:
[tex]\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643[/tex]
[tex]S_p=2.940[/tex]
[tex]t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751[/tex]
[tex]df=60+72-2=130[/tex]
[tex]p_v =P(t_{130}<-1.751) =0.0412[/tex]
Assuming a significance level of [tex]\alpha=0.05[/tex] we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane
Step-by-step explanation:
Data given
Our notation on this case :
[tex]n_1 =60[/tex] represent the sample size for people who used a self service
[tex]n_2 =72[/tex] represent the sample size for people who used a cashier
[tex]\bar X_1 =5.2[/tex] represent the sample mean for people who used a self service
[tex]\bar X_2 =6.1[/tex] represent the sample mean people who used a cashier
[tex]s_1=3.1[/tex] represent the sample standard deviation for people who used a self service
[tex]s_2=2.8[/tex] represent the sample standard deviation for people who used a cashier
Assumptions
When we have two independent samples from two normal distributions with equal variances we are assuming that
[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]
The statistic is given by:
[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]
And t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:
[tex]\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]
System of hypothesis
Null hypothesis: [tex]\mu_1 \geq \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 < \mu_2[/tex]
This system is equivalent to:
Null hypothesis: [tex]\mu_1 - \mu_2 \geq 0[/tex]
Alternative hypothesis: [tex]\mu_1 -\mu_2 < 0[/tex]
We can find the pooled variance:
[tex]\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643[/tex]
And the deviation would be just the square root of the variance:
[tex]S_p=2.940[/tex]
The statistic is given by:
[tex]t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751[/tex]
The degrees of freedom are given by:
[tex]df=60+72-2=130[/tex]
And now we can calculate the p value with:
[tex]p_v =P(t_{130}<-1.751) =0.0412[/tex]
Assuming a significance level of [tex]\alpha=0.05[/tex] we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane
