Question

6.224 exercise 2.153 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bills a group. in the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. the data on splitbill includes the cost of what each person ordered (in Israeli shekels)and the paymentethod(individual or split). some summary statistics are provided in table 6.20 and both distribution are reasonably well shaped. Use this information to test (at a 5% level) if there is evidence that the mean cost is higher when people spilt the bill. you may have done this test using randomization in exercise 4.118.
        table 6.20 cost of meals by payment type
        payment.      sample size.      mean.    std.dev
        individual.      24.                       37.29.    12.54
        split.                24.                       50.92.     14.33

Answer 1

Let $\mu_1$ be the the average.bill cost when the bill is paid individually.

Let $\mu_2$ be the the average bill cost when the bill is split..

Given:

For: $\bar{x}_1$ =37.29, s₁ = 12.54, n₁ = 24

For: $\bar{x}_2$ = 50.92, s₂ = 14.33, n₂ = 24

Since s₁/s₂ = 12.54 / 14.33 = 0.875 (it lies between 0.5 and 2) we used the pooled standard deviation

$S_p^2 = \frac{(n1-1)*s1^2+(n2-1)*s2^2}{n1+n2-2} = \frac{23*37.29^2+23*14.33^2}{24+24-2} = 181.3$

df = n1 + n2 - 2 = 24 + 24 - 2 = 46

The Hypothesis:

H0: $\mu_1$ = $\mu_2$

Ha: $\mu_1$ < $\mu_2$

This is a Left tailed test.

The Test Statistic:

$t = \frac{\bar{x}_1-\bar{x}_2}{\sqrt{S_p^2(\frac{1}{n1}+\frac{1}{n2})}} = \frac{37.29-50.92}{\sqrt{181.3*(\frac{1}{24}+\frac{1}{24})}} = -3.51$

t observed = -3.51

The p Value:    The p value (Left tail) for t = -3.51, df = 46,is; p value = 0.0005

The Critical Value:   The critical value (Left tail) at $\alpha$ = 0.05, df = 46, t critical = -1.679

The Decision Rule: If t observed is <- t critical Then Reject H0.

Also If the P value is < $\alpha$ , Then Reject H0

The Decision:    Since t observed (-3.51) is < t critical (-1.679), We Reject H0.

Also since P value (0.0005) is < $\alpha$ (0.05), We Reject H0.

The Conclusion:   There is sufficient evidence at the 95% significance level to conclude that the mean cost is higher when people split the bill.