Mathematics, 24.11.2020 03:40 brinatperez06

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 19% . Give 20 patrons of a restaurant a message on their bill warning them that tomorrow's weather will be bad and record the percentage tip they leave. Given are the tips as a percentage of the total bill.

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Answer from: pandadude123

The complete question is :

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 $H_0$ and $H_a$ 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,

$\bar x= \frac{\Sigma x}{n}$

$=\frac{18+19.1+19.2+...+19.4}{20}$

= 18.19

Now test : μ < 20 with normal distance

Hypothesis test:

$H_0=\mu \geq20$

$$H_a=\mu$

This is the lower tailed test.

$\bar x = 18.19$

$\sigma =2$

n = 20

Significant level, α = 0.05

Test statistics, $z^*=\frac{\bar x - \mu}{\sigma / \sqrt n }$

$=\frac{18.19 - 20}{2 / \sqrt{20}}$

≈ -4.05

P value is in direction of alternative hypothesis or $H_a$

Since, $$H_a=\mu$ , 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 $H_0$ if P value < α

Using excel functional normdist $\left(\frac{18.19-20}{2 / \sqrt{20}}, 0, 1 , \text{TRUE} \right )$ or TI's function normaldf $\left( 1E-99, \frac{18.19-20}{2 / \sqrt{20}}\right)$ 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.

Answer from: Quest

Where are the functions?

Answer from: Quest

I'm just guessing, but my answer would c. 0 to 300.

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