This would result in a biased sample because:
- the survey is surveying elementary kids about a new youth center
most of those kids are more likely than not to want a youth center, so most of them will naturally agree without a doubt for the need of a youth center
- Even the adults at the elementary school are more likely than not to agree with a new youth center because they are teachers and can even benefit from working at the youth center
- in final words the reason why this survey will be biased is because there is no variety in the participants being tested. It is basically just asking teachers she students their preference, excluding the rest of an entire community
please vote my answer brainliest. thanks!
Ni : Zn: Cu = 7:2:9 =7x : 2x : 9x
7x+2x+9x = 3.8
18x=3.8
x=3.8/18=1.9/9≈0.21
Ni: 7x=7*0.21=1.47 ≈1.5 kg
Zn: 2x=2*0.21=0.42≈0.4 kg
Cu: 9x=9*0.21=1.89 ≈1.9 kg
Check 1.5+0.4+1.9 =3.8 True
Answer:
The critical value is T = 1.895.
The 90% confidence interval for the mean repair cost for the washers is between $48.159 and $72.761
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 8 - 1 = 6
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 6 degrees of freedom(y-axis) and a confidence level of
. So we have T = 1.895, which is the critical value.
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 60.46 - 12.301 = $48.159
The upper end of the interval is the sample mean added to M. So it is 60.46 + 12.301 = $72.761
The 90% confidence interval for the mean repair cost for the washers is between $48.159 and $72.761