Frank just completed two tests. On his biology test, he earned a 55 while he earned a 70 on his chemistry test. The class averag
2 answers:
Since we cannot be 100% sure which test Frank did better on, we need to find a way to make the comparison doable. This can be done by converting Frank's score to the standard score (or also known as the z-score).
The z-score is calculated through the following formula:
where X is Frank's original score.
Converting Frank's scores to a standard score accounting for both the mean and standard deviation of the population who took the test will make the comparison possible.
We get the following z-scores after conversion:
If you read up more on the concept of z-score, you may know that this score just means the amount of standard deviations a data falls with respect to the mean. Comparing Frank's two scores, we can see that they are both below the mean. However, it can be seen that Frank's z-score for his chemistry test is lower than his z-score for his biology test.
This suggests that Frank scored relatively poorer in chemistry than he did in biology.
ANSWER: Frank did better in his biology test.
The z-score is calculated through the following formula:
Converting Frank's scores to a standard score accounting for both the mean and standard deviation of the population who took the test will make the comparison possible.
We get the following z-scores after conversion:
If you read up more on the concept of z-score, you may know that this score just means the amount of standard deviations a data falls with respect to the mean. Comparing Frank's two scores, we can see that they are both below the mean. However, it can be seen that Frank's z-score for his chemistry test is lower than his z-score for his biology test.
This suggests that Frank scored relatively poorer in chemistry than he did in biology.
ANSWER: Frank did better in his biology test.
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Answer:
a) σ/√n= 1.43 min
c) Margin of error 2.8028min
d) [30.1972; 35.8028]min
e) n=62 customers
Step-by-step explanation:
Hello!
The variable of interest is
X: Time a customer stays at a restaurant. (min)
A sample of 49 lunch customers was taken at a restaurant obtaining
X[bar]= 33 mi
The population standard deviation is known to be δ= 10min
a) and b)
There is no information about the distribution of the population, but we know that if the sample is large enough, n≥30, we can apply the central limit theorem and approximate the distribution of the sample mean to normal:
X[bar]≈N(μ;σ²/n)
Where μ is the population mean and σ²/n is the population variance of the sampling distribution.
The standard deviation of the mean is the square root of its variance:
√(σ²/n)= σ/√n= 10/√49= 10/7= 1.428≅ 1.43min
c)
The CI for the population mean has the general structure "Point estimator" ± "Margin of error"
Considering that we approximated the sampling distribution to normal and the standard deviation is known, the statistic to use to estimate the population mean is Z= (X[bar]-μ)/(σ/√n)≈N(0;1)
The formula for the interval is:
[X[bar]±*(σ/√n)]
The margin of error of the 95% interval is:
d= *(σ/√n)= 1.96* 1.43= 2.8028
d)
[X[bar]±*(σ/√n)]
[33±2.8028]
[30.1972; 35.8028]min
Using a confidence level of 95% you'd expect that the interval [30.1972; 35.8028]min contains the true average of time the customers spend at the restaurant.
e)
Considering the margin of error d=2.5min and the confidence level 95% you have to calculate the corresponding sample size to estimate the population mean. To do so you have to clear the value of n from the expression:
d= *(σ/√n)
= σ/√n
√n*()= σ
√n= σ* ()
n=( σ* ())²
n= (10*)²= 61.47≅ 62 customers
I hope this helps!