ASAP PLS ILL GIVE BRAINLIEST
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Answer:
We conclude that the population mean is different from 10.5.
Step-by-step explanation:
We are given that a random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 11 and the sample standard deviation is 2.
<em>We have to test the claim that the population mean is 10.5.</em>
Let, NULL HYPOTHESIS, :
= 10.5 {means that the population mean is 10.5}
ALTERNATE HYPOTHESIS, :
10.5 {means that the population mean is different from 10.5}
The test statistics that will be used here is One-sample t-test;
T.S. = ~
where, = sample mean = 11
s = sample standard deviation = 2
= population mean
n = sample of values = 16
So, <u>test statistics</u> = ~
= 1
<em>Now, at 0.05 significance level, t table gives a critical value of 2.131 at 15 degree of freedom. Since our test statistics is way less than the critical value of t so we have insufficient evidence to reject null hypothesis as it will not fall in the rejection region.</em>
Therefore, we conclude that the population mean is different from 10.5.
So firstly, we have to find the LCD, or lowest common denominator, of 9 and 7. To do this, list the multiples of 9 and 7 and the lowest multiple they share is going to be your LCD. In this case, the LCD of 9 and 7 is 63. Multiply x^2/9 by 7/7 and 2y/7 by 9/9:
Next, add the numerators together, and your answer will be: