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People tend to be more generous after receiving good news. Are they less generous after receiving bad news? The average tip left by adult Americans is 20%. Give 20 patrons of a restaurant a message on their bill warning them that tomorrow’s weather will be bad and record the tip percentage they leave. Table 1 contains the tips as a percentage of the total bill.
Table : 18.0 19.1 19.2 18.8 18.4 19.0 18.5 16.1 16.8 18.2 14.0 17.0 13.6 17.5 20.0 20.2 18.8 18.0 23.2 19.4
Suppose that tip percentage is Normal with ? = 2 and assume that the patrons in this study are a random sample of all patrons of this restaurant. Is there good evidence that the mean tip percentage for all patrons of this restaurant is less than 20 when they receive a message warning them of tomorrow’s bad weather? State [tex]$H_0$[/tex] and [tex]$H_a$[/tex] and carry out a significance test. Use significance level 0.05.
Solution:
The given data is :
18.0, 19.1, 19.2, 18.8, 18.4, 19.0, 18.5, 16.1, 16.8, 18.2, 14.0, 17.0, 13.6, 17.5, 20.0, 20.2, 18.8, 18.0, 23.2, 19.4
Therefore, sample mean,
[tex]$\bar x= \frac{\Sigma x}{n}$[/tex]
[tex]$=\frac{18+19.1+19.2+...+19.4}{20}$[/tex]
= 18.19
Now test : μ < 20 with normal distance
Hypothesis test:
[tex]$H_0=\mu \geq20$[/tex]
[tex]$H_a=\mu <20$[/tex]
This is the lower tailed test.
[tex]$\bar x = 18.19$[/tex]
[tex]$\sigma =2$[/tex]
n = 20
Significant level, α = 0.05
Test statistics, [tex]$z^*=\frac{\bar x - \mu}{\sigma / \sqrt n }$[/tex]
[tex]$=\frac{18.19 - 20}{2 / \sqrt{20}}$[/tex]
≈ -4.05
P value is in direction of alternative hypothesis or [tex]$H_a$[/tex]
Since, [tex]$H_a=\mu <20$[/tex] , P value = P(Z<-4.05)
P value is the area to the left of the test statistics = -4.05 (from the figure)
P value = P(Z< -4.05) = 0.000 (from z table)
P value = 00000
Rejection rule : Reject [tex]$H_0$[/tex] if P value < α
Using excel functional normdist [tex]$\left(\frac{18.19-20}{2 / \sqrt{20}}, 0, 1 , \text{TRUE} \right )$[/tex] or TI's function normaldf [tex]$\left( 1E-99, \frac{18.19-20}{2 / \sqrt{20}}\right)$[/tex] answer is 0.0000259078
Decision : Since 0.0000 < 0.05, we reject the null hypothesis
Hence at 5% significance level, we have sufficient evidence to conduct that μ is less than 20.
At the 5% significance level, we have shown that μ is less than 20.
What are test statistics?
A test statistic is defined in such a way as to the quantity, within observed data that would distinguish the null from the alternative hypothesis
[tex]{\displaystyle z={\frac {{\overline {x}}-\mu _{0}}{({\sigma }/{\sqrt {n}})}}}[/tex]
The given data is
18.0, 19.1, 19.2, 18.8, 18.4, 19.0, 18.5, 16.1, 16.8, 18.2, 14.0, 17.0, 13.6, 17.5, 20.0, 20.2, 18.8, 18.0, 23.2, 19.4
Mean = Sum of the terms / number of terms
= [tex]\frac{18.0 + 19.1 + 19.2, +........+23.2 1+ 9.4}{20}[/tex]
= 18.19
Now with a normal distance test, we get
[tex]\rm H_{0} = \mu \geq 20\\\rm H_{\alpha } = \mu < 20[/tex]
lower tailed test
Mean = 18.19
[tex]\sigma = 2[/tex]
n = 20
Test statistics, [tex]{\displaystyle z={\frac {{\overline {x}}-\mu _{0}}{({\sigma }/{\sqrt {n}})}}}[/tex]
z = [tex]\frac{18.19-20}{2/\sqrt{20} }[/tex]
= -4.05
Significant level, α = 0.05
P-value is in direction of an alternative hypothesis or
Since, P-value = P(Z<-4.05)
= 0.000 (from z table)
P value = 00000
By Rejection rule, Reject [tex]\rm H_{0}[/tex] if P value < α
Using excel functional normdist, the answer is 0.0000259078
Since 0.0000 < 0.05, we reject the null hypothesis
Therefore, at the 5% significance level, we have shown that μ is less than 20.
Learn more about Test statistics;
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