5x + 6 = 5x - 4 does it have an unique solution
2 answers:
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<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).
- The lower limit is 0.5.
- The upper limit is 0.7.
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the z-score that has a p-value of .
60 out of 100 scores are passing scores, hence
95% confidence level
So , z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).
- The lower limit is 0.5.
- The upper limit is 0.7.
A similar problem is given at brainly.com/question/16807970
Answer:
We conclude that this is an unusually high number of faulty modems.
Step-by-step explanation:
We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.
The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.
Let p = <em><u>population proportion</u></em>.
So, Null Hypothesis, : p = 0.013 {means that this is an unusually 0.013 proportion of faulty modems}
Alternate Hypothesis, : p > 0.013 {means that this is an unusually high number of faulty modems}
The test statistics that would be used here <u>One-sample z-test</u> for proportions;
T.S. = ~ N(0,1)
where, = sample proportion faulty modems=
= 0.027
n = sample of modems = 367
So, <u><em>the test statistics</em></u> =
= 2.367
The value of z-test statistics is 2.367.
Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>
Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.
Therefore, we conclude that this is an unusually high number of faulty modems.