Mr. Malloy wants to make sure Amo and Javier get the best possible grade after five tests and can either give a grade based on t
he median or the mean.
Test
Number
Amo’s Scores
Javier’s Scores
1
97
68
2
92
97
3
89
65
4
55
92
5
90
65
What measure of center should Mr. Malloy use for Amo and for Javier for the best possible grade?
The teacher should use the mean for both Amo and Javier.
The teacher should use the median for both Amo and Javier.
The teacher should use the median for Amo and the mean for Javier.
The teacher should use the mean for Amo and the median for Javier.
Test
Number
Amo’s Scores
Javier’s Scores
1
97
68
2
92
97
3
89
65
4
55
92
5
90
65
What measure of center should Mr. Malloy use for Amo and for Javier for the best possible grade?
The teacher should use the mean for both Amo and Javier.
The teacher should use the median for both Amo and Javier.
The teacher should use the median for Amo and the mean for Javier.
The teacher should use the mean for Amo and the median for Javier.
2 answers:
The data has been properly arranged in tabular form and is shown below in the image.
First we need to find the mean and median of scores of both students.
1) For Amo:
Mean =
Median = Middle Value when data is arranged in ascending order = 90
2) For Javier:
Mean =
Median = Middle Value when data is arranged in ascending order = 92
For both the students, value of Median is larger then the mean. So in order to give the best possible grade Mr. Malloy should use the median score for both students.
First we need to find the mean and median of scores of both students.
1) For Amo:
Mean =
Median = Middle Value when data is arranged in ascending order = 90
2) For Javier:
Mean =
Median = Middle Value when data is arranged in ascending order = 92
For both the students, value of Median is larger then the mean. So in order to give the best possible grade Mr. Malloy should use the median score for both students.
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Answer:
a. Descriptive statistics for the sample.
n= 9
mean X[bar]= 20.36
median Me= 20.30
min: 19.70
max: 21.40
standad deviation S= 0.61
b. 95% Confidence interval [19.88; 20.83]
c. Decision: Support H₀
Step-by-step explanation:
Hello!
I don't have excel so I cannot install PHstat, I've used a statistic software called Infostat for the calculations.
There was a random sample of 9 bottles of shampoo taken to test if the process is operating correctly. If it is, the bottles should weight on average 20 ounces (this would be our study parameter)
The study variable is X: the weight of a bottle of shampoo.
X~N(μ;σ²)
a.
Descriptive statistics for the sample.
n= 9
mean X[bar]= 20.36
median Me= 20.30
min: 19.70
max: 21.40
standad deviation S= 0.61
b. 95% Confidence interval.
Since the variable has a normal distribution and the population variance is unknown, the statistic to use to construct this interval is the Students t:
X[bar] ± * (S/√n)
20.36 ± * (0.61/√9)
[19.88; 20.83]
c. You have to test the hypothesis that the production is operating correctly, if it is so, then:
H₀: μ = 20
H₁: μ ≠ 20
α: 0.05
The statistic value is t= 1.74
p-value: 0.1198
When you use the p-value to decide on a statistical hypothesis, you should always contrast it with the level of significance using the following rule:
If p-value ≤ α, then you reject the null hypothesis.
If p-value > α, then you do not reject the null hypothesis.
Since the p-value is greater than the significance level, you do not reject the null hypothesis. This means that the average weight of the shampoo bottles is 20 ounces, i.e. the process is operating correctly.
I hope it helps!
Two one hundred is 200.
15 tens is 150.
Six ones is the same.
15 tens is 150.
Six ones is the same.