Step-by-step explanation:
only 10 and 12. luck to you
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1 answer:
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The correct statements are as follows;
42% of the students prefer a concert over a movie.
A greater ratio of 11th graders prefer a concert to a movie
The difference in the percent of students who prefer a movie over a concert is 16%
Given that
A random sample of 184 students at South High School was surveyed on whether they would prefer going to a movie or a concert.
10th 11th 12th total
Movie .18 .16 .24 .58
Concert .15 .18 .09 .42
Total .33 .34 .33 1
According to the question
The marginal relative frequency for 10th graders is 0.18.
This is an incomplete statement.
The marginal frequency is a number, not a percent.
42% of the students prefer a concert over a movie.
42% preferred a convert to the 58% who preferred a movie.
This statement is True.
A greater ratio of 11th graders prefers a concert to a movie.
18/100 concert to 16/100 for the movie for 11th graders.
More 12th graders were surveyed than any other grade level.
There were more 11th graders surveyed (34%) to (33%) for 12 graders False
The difference in the percent of students who prefer a movie over a concert is 16%.
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Answer:
We conclude that the true average percentage of organic matter in such soil is different from 3%.
Step-by-step explanation:
We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.
Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.
<em>Let </em><em> = true average percentage of organic matter in such soil</em>
SO, <u>Null Hypothesis</u>, :
= 3% {means that the true average percentage of organic matter in such soil is equal to 3%}
<u>Alternate Hypothesis</u>, :
3% {means that the true average percentage of organic matter in such soil is different than 3%}
The test statistics that will be used here is <u>One-sample t test statistics</u> because we don't know about the population standard deviation;
T.S. = ~
where, = sample mean amount of organic matter = 2.481%
s = sample standard deviation = 1.616%
n = sample of soil specimens = 30
So, <u><em>test statistics</em></u> = ~
= -1.759
<u></u>
<u>Now, P-value of the test statistics is given by;</u>
P-value = P( > -1.759) = <u>0.046</u> or 4.6%
- If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.
- If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.
<em>Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.</em>
Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.