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tatuchka [14]
3 years ago
11

5) In a certain supermarket, a sample of 60 customers who used a self-service checkout lane averaged 5.2 minutes of checkout tim

e, with a standard deviation of 3.1 minutes. A sample of 72 customers who used a cashier averaged 6.1 minutes with a standard deviation of 2.8 minutes. Suppose it is reasonable to assume that the two variance of the two groups are the same. Can you conclude that the mean checkout time is less for people who use the self-service lane
Mathematics
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

S_p=2.940

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

df=60+72-2=130

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 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 :

n_1 =60 represent the sample size for people who used a self service

n_2 =72 represent the sample size for people who used a cashier

\bar X_1 =5.2 represent the sample mean for people who used a self service

\bar X_2 =6.1 represent the sample mean people who used a cashier

s_1=3.1 represent the sample standard deviation for people who used a self service

s_2=2.8 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  

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

And t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

System of hypothesis

Null hypothesis: \mu_1 \geq \mu_2

Alternative hypothesis: \mu_1 < \mu_2

This system is equivalent to:

Null hypothesis: \mu_1 - \mu_2 \geq 0

Alternative hypothesis: \mu_1 -\mu_2 < 0

We can find the pooled variance:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

And the deviation would be just the square root of the variance:

S_p=2.940

The statistic is given by:

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

The degrees of freedom are given by:

df=60+72-2=130

And now we can calculate the p value with:

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 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

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