Charlie withdraws cash from an ATM that is not his own bank’s 3 times a month. He pays $1.50 per transaction. If this pattern is
2 answers:
You might be interested in
Answer:
a)
The degrees of freedom are given by:
The p value for this case would be given by:
Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425
b) For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:
Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425
Step-by-step explanation:
Information given
represent the sample mean for the cubic feets of households
represent the population standard deviation
sample size
represent the value to verify
represent the significance level
t would represent the statistic
represent the p value
Part a
We want to test that the mean consumption of water per household has decreased due to the campaign by the city council, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
Since we don't know the deviation the statistic is given by:
(1)
The statistic is given by:
The degrees of freedom are given by:
The p value for this case would be given by:
Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425
Part b
For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:
Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425
The box-and-whisker plot is attached.
We first order the data from least to greatest:
67,68,68,69,70,72,73,74,75,75,76,78,79,79,80,83,84,85,85,88
Now we find the median. Since there are 20 data values, the median is between 75 and 76:
(75+76)/2 = 151/2 = 75.5
Now we find the Lower Quartile (LQ). We do this by finding the median of the lower half of data (from the median down). There are 10 values here; the median is between 70 and 72:
(70+72)/2 = 142/2 = 71
Find the Upper Quartile (UQ). Do this by finding the median of the upper half of data (from the median up). There are 10 values here; the median is between 80 and 83:
(80+83)/2 = 163/2 = 81.5
The middle bar of the box is over the median. The left hand side of the box is over LQ, and the right hand side of the box is over UQ. Now we draw a whisker from the box to the highest value, 88, and from the box to the lowest value, 67.
We first order the data from least to greatest:
67,68,68,69,70,72,73,74,75,75,76,78,79,79,80,83,84,85,85,88
Now we find the median. Since there are 20 data values, the median is between 75 and 76:
(75+76)/2 = 151/2 = 75.5
Now we find the Lower Quartile (LQ). We do this by finding the median of the lower half of data (from the median down). There are 10 values here; the median is between 70 and 72:
(70+72)/2 = 142/2 = 71
Find the Upper Quartile (UQ). Do this by finding the median of the upper half of data (from the median up). There are 10 values here; the median is between 80 and 83:
(80+83)/2 = 163/2 = 81.5
The middle bar of the box is over the median. The left hand side of the box is over LQ, and the right hand side of the box is over UQ. Now we draw a whisker from the box to the highest value, 88, and from the box to the lowest value, 67.