Answer:
We are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%
Step-by-step explanation:
We are given that in a group of randomly selected adults, 160 identified themselves as executives.
n = 160
Also we are given that 42 of executives preferred trucks.
So the proportion of executives who prefer trucks is given by
p = 42/160
p = 0.2625
We are asked to find the 95% confidence interval for the percent of executives who prefer trucks.
We can use normal distribution for this problem if the following conditions are satisfied.
n×p ≥ 10
160×0.2625 ≥ 10
42 ≥ 10 (satisfied)
n×(1 - p) ≥ 10
160×(1 - 0.2625) ≥ 10
118 ≥ 10 (satisfied)
The required confidence interval is given by
Where p is the proportion of executives who prefer trucks, n is the number of executives and z is the z-score corresponding to the confidence level of 95%.
Form the z-table, the z-score corresponding to the confidence level of 95% is 1.96
Therefore, we are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%
Answer:
Check explanation.
Step-by-step explanation:
slope-intercept form: y=mx+b
9x+5y=23
5y=-9x+23
y=-9/5x+23/5
m=slope: -9/5
b=y-intercept: 23/5
The first one is not a function due to the rule that there can not be more than one x, such as there is a repeated number 1 , 2 , 1 , 4 There should not be two
x = 1 terms.
Same the same rule applies to the second option as well as the last.
Your answer is C. or the third option
x= 6, 5, 4, 1
y=6, 4, 6, 2
Hope this Helps
Does it tell you what (x) stands for
The required type of p-value would she want to obtain is small p-value.
<h3>What is hypothesis test?</h3>
Hypothesis testing is a type of measurable surmising that utilizes information from an example to make inferences about a populace boundary or a populace probability distribution. Initial, a speculative supposition that is made about the boundary or dispersion.
<h3>According to question:</h3>
P-value shows that the outcomes are genuinely huge.
P-value in Measurements assists with performing speculation test and it assists with deciding the meaning of the outcomes.
A little P-value commonly (P<0.05) demonstrates solid proof against the invalid speculation in this way, we reject the invalid speculation, on opposite when P-esteem is enormous we can't dismiss the invalid speculation.
Thus, required answer is small p-value.
To know more about p-value visit:
brainly.com/question/14790912
#SPJ4
