You didn't show us the choices.
The graph of that function would be centered on the y-axis.
A) The mean of set 1 is 53,800, while the mean of set 2 is 38,600. The median of set 1 is 42,000, while the median of set 2 is 28,000.
B) The median is a more accurate estimate, because each set has 1 particularly large outlier which heavily skews the mean. In each set, the mean is actually higher than 4 out of the 5 data points, which is not very accurate in describing how much the typical car costs.
Answer:
Step-by-step explanation:
Hello!
The chemistry instructor tested the hypothesis that the proportion of students that passed the introductory chemistry class is better with an embedded. If the known proportion for this population is 65%, the tested hypothesis is:
H₀: p=0.65
H₁: p>0.65
The calculated statistic is Z=2.52 and the associated p-value: 0.0059
Remember:
The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
In this case:
P(Z≥2.52)=0.0059
There is no significance level, the most common one is α: 0.05 so I'll use it as an example.
To make a decision using the p-value you have to compare it to the α.
If p- value>α then you support the null hypothesis (In this case, you can say that there is no change in the proportion of students that passed the introductory chemistry class with an embedded tutor.)
If p-value≤α your decision will be to reject the null hypothesis (In this case, there is significant evidence to say that there is an improvement in the success rate of the introductory chemistry class with an embedded tutor?
Since the p-value:0.0059 is less than the significance level 0.05, you will decide to reject the null hypothesis.
I hope you have a SUPER day!
3.5x8 = 28 grams have a good day
