Waiting in line. A quality - control manager at an amusement park feels that the amount of time that people spend waiting in line for the American Eagle roller coaster is too long. To determinate if a new loading/unloading procedure if efective in reducing wait time in line, he measured the amount of time (in minutes) people are waiting in line for 7 days. After implementing the new procedure, he again measures the amount of time in minutes and people are waiting in line 7 days and obtains the following data.

Wait time before new procedure
Day
Mon Tues Wed Thurs Fri Sat Sat Sun Sun
11.6 25.9 20.0 38.2 57.3 32.1 81.8 57.1 62.8

Wait time after new procedure
10.7 28.3 19.2 35.9 59.2 31.8 75.3 54.9 62.0

test the claim that the new loading/unloading procedure is effective in reducing wait time (H0: µd=0 and H1: µd<0)at α=.05 level of significance. Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers (use the classical approach and the p-value approach).

No

explanation:

given:

n=9

=0.05

see the attachment

Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

d=0.9-2.4+0.8+...+6.5+2.2+0.8/9

 =1.0556

The variance is the sum of squared deviations from the mean divided by n-1. The standard deviation is the square root of the variance. Determine the sample standard deviation of the differences:  

s_d=√(0.9-1.0556)^2+...+(0.8-1.0556)^2/9-1

     =2.6

CLASSICAL APPROACH :

Given claim: new procedure reduces > 0  

The claim is either the null hypothesis or the alternative hypothesis The null hypothesis and the alternative hypothesis state the opposite of each other The null hypothesis needs to contain an equality  

Determine the value of the test statistic

t=d-/s_d/√n

=1.220

Determine the critical value from the Student T distribution table in the appendix in the row with d_f = n- 1 = 9-1 = 8 and in the column with = 0.05  t =1.860  

The rejection region then contains all values larger than 1.860  

If the value of the test statistic is within the failed region, then the null hypothesis is failed

1.220 < 1.860 failed H_0

There is not sufficient evidence to support the claim that the new loading/unloading procedure is effective in reducing the wait time.  

P VALUE APPROACH:

Given claim: new procedure reduces  > 0  

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.  

Determine the value of the test statistic:  

 t=d-/s_d/√n

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. The P-value is the number (or interval) in the column title of the Students T distribution in the appendix containing the t-value in the row d_f = n-1 = 9-1 = 8

0.10 < P < 0.15

If the P-value is less than the significance level, reject the null hypothesis.  

P > 0.05   failed H_0

There is not sufficient evidence to support the claim that the new loading unloading procedure is effective in reducing the wait time.  

3.4 pizzas

step-by-step explanation:

given is a scatter plot showing number of pizzas sold during weeks for different issues of coupon

y represents number of pizzas sold and x number of coupons issued

y is marked verticaly and x horizontally

the trend line is given as

y =3.4x+43

this says that even when no coupon is issued pizzas sold will be 43

and rate of change of pizza sold for 1 unit of coupon = slope = 3.4

i.e. when x increases by 1 unit y increases by 3.4 units

hence for each additional coupon issued 3.4 pizzas more will be sold

the aswere is the one with all 8's and the second one is with the 10,8 and 6

step-by-step explanation: